Why Did My NPN/PNP Voltage Regulator Experiment Explode?
You already have an unregulated DC supply. As you say, built from a bridge and some capacitors. Apparently, you have a center-tap on your transformer secondary, too. So you have a ground, too, and \$\pm\:53\:\text{V}\$ measured with your meter for the two other rails. I'll assume that this is probably unloaded, so you probably will have less than that when loaded. How much less is anyone's guess, as it depends a lot on the loading, your toroid design, the capacitors, and other factors. But less, for sure.
I gather you are trying to learn about how to design your own \$\pm\: 15\:\text{V}\$ supply for use with opamps. So you aren't necessarily just wanting to buy a nice supply (they are cheap, these days.) And since this is about learning, it's going to be a linear design and not a switcher. So your power supply will be generally inefficient, power-wise. But you are fine with that.
Perhaps I'm projecting, but I think this is a good idea to start with. It's modest enough that you have every reason to succeed. But there is enough to learn about that it's worth struggling for, too. I think my very first learning experience, where I really learned a few things well, was in trying to design my own power supply like this. At the time, then, I pretty much didn't have a choice. Existing lab supplies were unobtainable for a young teenager. And there was no set of cheap ebay suppliers for fancy switchers based on ICs, either. So I had to do it myself or go without. And faced with that, one learns or one does without.
Your approach is perhaps a little too much like a sink/source output driver used in everything from opamps to audio amplifiers. You could take the approach you are taking, but you'd have to make two of them -- one for \$+15\:\text{V}\$ and one for \$-15\:\text{V}\$. And they are even less efficient, as they can each source from your (+) rail and sink to your (-) rail, and you need to run them in class-AB. You really only need to source from (+) to make the \$+15\:\text{V}\$ rail and to sink to (-) to make the \$-15\:\text{V}\$ rail.
Just as a side note, it may be a good idea to include a pair of bleeder resistors to your existing capacitor bank at the output of your bridge. Something to get rid of the stored charge if you turn things off. Some \$\tfrac{1}{2}\:\text{W}\$, \$10\:\text{k}\Omega\$ resistors? That would only present a \$5\:\text{mA}\$ load, when running.
While you are considering that idea, consider also trying to load down your existing unregulated supply to measure what it does under load. I'd try something like a \$\ge 5\:\text{W}\$, \$1\:\text{k}\Omega\$ resistor to get an idea about a \$50\:\text{mA}\$ load, measuring the voltage with that load present. I'd then try something like a \$\ge 10\:\text{W}\$, \$270\:\Omega\$ resistor to see what happens when I get near \$200\:\text{mA}\$ load. This will test your entire unregulated system and give you an idea about its limitations. Those values were picked at random. If you already know the limitations of your toroid, then try out two different resistor values that hit the maximum load you expect to support and another one to hit perhaps 30% of the maximum load. And just take note of the voltage values measured. It helps to have an idea about your unregulated rail when loaded down a bit.
I'd recommend that you start by focusing on just one side, say creating the \$+15\:\text{V}\$ regulated supply rail from your unregulated (+) rail. You need to consider whether or not you want any current limits, too. I think it would be safer to include them. But that's your decision. It's not hard to include something for that, though. And, just personally, I'd probably want to be able to go to \$+12\:\text{V}\$, too. So perhaps a variable output supply that works over some modest range of output voltages?
You have lots of headroom! This means you can use an NPN emitter follower, a Darlington follower, or just about any configuration you want to have. Things are not tight, so you have room for control structures. Lots of room. The downside is, of course, that you have to dissipate and that your voltage rails are enough to make you have to check datasheets to stay within safe operating parameters for devices.
Finally, you can probably accept having to separately set the two voltage rail values, independently. Some power supplies are designed to provide tracking so that if you set the regulated \$+\text{V}\$ supply to \$+15\:\text{V}\$ then your regulated \$-\text{V}\$ supply will track that and provide \$-15\:\text{V}\$. But you can live without that, for now, I suspect.
If you write up a separate question, or clarify this one better, I may get you started with three or four different discrete (non-IC) topologies to consider analyzing on your own and building. But, for example, I have no idea what kind of current compliance you want to have. And it would help to know what voltage you measure when your unregulated supply is loaded down to the maximum current compliance you want to support (using a high wattage resistor and then taking a moment to measure the voltage with a voltmeter before it gets too hot.) And it would help still more to know if you do want a variable voltage over a range (what range, exactly?) and, if you just want a fixed voltage, how much initial accuracy do you feel you need? And I'd like to know if this is strictly for an opamp supply (suggesting a lower current compliance) or if you will want to use this to actually supply higher currents at still lower voltages, for some projects. Finally, it would be nice to know what BJTs you have, or are willing to get.
EDIT: So. Something simple, not very much current compliance of only \$5\:\text{mA}\$. Let's first focus on the (+) rail side... could go either with NPN or PNP for the pass transistor. It's more a matter of how you want to control it. Do you want to siphon away current from a source, or pull out current as needed? Hmm. Let's try this -- emphasis on simple.



simulate this circuit &ndash; Schematic created using CircuitLab
I've written down some design notes on the schematic. The resistor values are standard ones, so the actual output voltage will be a little off. But it should be close. Here's the logic.
I started out using \$Q_1\$ as an emitter follower topology. It's emitter targets \$15\:\text{V}\$. So I wrote down &quot;15V @ 5mA&quot; there. I initially estimated a useful \$\beta_{Q1}=50\$ and computed \$I_{B_{Q1}}=100\:\mu\text{A}\$ and estimated (from memory only) \$V_{BE_{Q1}}=750\:\text{mV}\$. From this, I decided I wanted \$5\times\$ as much from the unregulated supply, so I set \$R_1=\frac{53V-15V-750\:\text{mV}}{500\:\mu\text{A}}=74.5\:\text{k}\Omega \approx 75\:\text{k}\Omega\$. This means that I'll need to pull away between \$400-500\:\mu\text{A}\$ from \$R_1\$ to control \$Q_1\$'s behavior at the output. That's a small enough range, \$450\:\mu\text{A}\pm 50\:\mu\text{A}\$, that variations in a simple circuit won't be too sensitive. Oh, and I chose the BC546, which has a \$V_{CEO}=65\:\text{V}\$. (Could use a 2N5551 for \$V_{CEO}=150\:\text{V}\$.)
I decided to use another NPN down below, with its base nailed to a resistor divider, to pull that current. \$Q_2\$'s collector is nailed to a voltage, so no Early Effect. Fine. Dissipation in \$Q_2\$ is under \$10\:\text{mW}\$, so no problem. (You already know there may be a problem in \$Q_1\$.) A diode and capacitor provides a semi-stable voltage reference, as it is fed a relatively stable \$450\:\mu\text{A}\pm\:50\:\mu\text{A}\$ current. I estimated \$\beta_{Q2}=50\$ (again) and computed \$I_{B_{Q2}}=10\:\mu\text{A}\$ and estimated (from memory only) \$V_{BE_{Q1}}=650\:\text{mV}\$. I also know that the 1N4148 does about \$550\:\text{mV}\$ running at \$500\:\mu\text{A}\$ current. So this told me that the divider node should be guessed at \$1.2\:\text{V}\$. I wrote that down, too.
I chose to make the divider current at least \$10\times\$ the max required base current for \$Q_2\$. One of the problems with this circuit is going to be ambient temperatures, as these affect the base-emitter junction of \$Q_2\$ (and \$D_1\$, too) and this affects our divider point and pretty much everything else. But adding \$D_2\$ and \$D_3\$ in the divider helps here. It provides two more temperature dependent junctions that will track the other two over temperature. The remaining problem being \$R_3\$ and the differing current densities.
\$D_2\$ and \$D_3\$ are running with about \$\tfrac{1}{5}\$ of the current density of \$D_1\$ and \$Q_2\$. I happen to remember that a 1N4148 presents about \$\Delta V \approx 100\:\text{mV}\$ per decade change in current density, so I guess that \$\Delta V = 100\:\text{mV}\cdot \log_{10}\left(\tfrac{1}{5}\right) \approx -70\:\text{mV}\$ per diode for those two. So this means that to reach \$1.2\:\text{V}\$ at the divider, \$R_3=\frac{1.2V - 2\cdot\left(550\:\text{mV}-70\:\text{mV}\right)}{87\:\mu\text{A}}\approx 2.7\:\text{k}\Omega\$ (I used \$87\:\mu\text{A}\$ as the mid-point current value.) So that sets \$R_3\$, at a guess.
I added a speed up cap across divider resistor \$R_2\$ so that short-term load variations might more immediately drive \$Q_2\$. (If the \$15\:\text{V}\$ regulated rail suddenly jumps upward, then \$C_3\$ will pull up immediately on the base of \$Q_2\$ making it pull away more of the drive current going to \$Q_1\$, countering the rise. Similarly, in the other direction, too.)
You should be able to pony up the (-) regulated rail, I think. And keep in mind that you do not want to load this thing down too much! You will definitely cause that poor little TO-92 serious problems. It's dissipating \$5\:\text{mA}\cdot\left(53\:\text{V}-15\:\text{V}\right)\approx 200\:\text{mW}\$ and the package has \$\tfrac{200 ^{\circ}K}{W}\$, so this works out to about \$+40^{\circ}C\$ over ambient, already. You can see just how quickly this thing will heat up if you run much more current through it. You may be able to get away with \$10\:\text{mA}\$, but not much more.
OVERVIEW NOTE: Now that you can see one person's process (other, more experienced designers will apply still more knowledge than I applied), let's take a moment to view this from a distant perspective.
The circuit boils down to:

A pass transistor (\$Q_1\$) which is supposed to stand-off about \$40\:\text{V}\$ between the unregulated (+) rail and the desired \$15\:\text{V}\$ rail. This pass transistor will need a source of base current so that it can be kept in its active region. It is also arranged into an emitter-follower configuration, so that moving its base voltage around moves its emitter around in roughly 1:1 (voltage gain from base to emitter is \$\approx 1\$.)
We can solve all of the needs in (1) above by using a simple resistor (\$R_1\$) to the unregulated (+) rail. This not only can provide the needed base current, but it also makes it very easy to control the base voltage of \$Q_1\$, by just pulling more or less current through it. For design purposes, we do not want variations in \$Q_1\$'s base current to seriously impact the current stream we are also using to control the voltage at the base of \$Q_1\$. So we need to make this stream of current large, by comparison. Larger is better, and perhaps by default we might choose a factor of \$10\times\$. But we are also constrained by the fact that this is a \$5\:\text{mA}\$ power supply. So we might want to use something that is about \$\tfrac{1}{10}\$th of \$5\:\text{mA}\$ to keep it modest. This means something from \$10\cdot 100\:\mu\text{A}=1\:\text{mA}\$ on the one side to about \$\tfrac{5mA}{10}=500\:\mu\text{A}\$ on the other side. I decided to use the smaller value, since this is just a simple regulator and I can accept a slightly less stiff base source.
Something to control the current being pulled through \$R_1\$, based upon a voltage comparison of some kind. It turns out that a BJT is okay for something like this. (More BJTs would be better, as in an opamp, but one is sufficient here.) It has a collector current that depends upon the voltage difference between its base and emitter. So it compares its' base and emitter and adjusts a current on that basis! Practically made in heaven for this, yes? So we now stick a new BJT (\$Q_2\$) with its collector tied up to \$R_1\$ and the base of \$Q_1\$.
We need a reference voltage. Could use a real reference, like a zener or a more sophisticated IC device, but this is a simple design. Well, a diode with a fixed current density is a voltage reference. (Excepting temperature.) And guess what? We just happen to have a current we can use that is relatively stable! The very current we are using to adjust \$Q_1\$'s base voltage through \$R_1\$. So now, \$R_1\$ provides three services for us -- it provides base current to \$Q_1\$, allows us to control \$Q_1\$'s base by adjusting the current through it, and now that very same current can be used to stabilize the voltage of a voltage reference diode. All we do is stick that diode into the emitter of \$Q_2\$. And add a small capacitor across it o kill high frequency noise there. It's nice when things do multiple duties for you.
We have our current control collector, a voltage reference at the emitter, and now all we need to provide is a comparison voltage, derived from the output voltage, at the base of \$Q_2\$. It's important that if this comparison increases (the output voltage appears to increase for some unknown reason), that we will pull more current through \$R_1\$ to force the base voltage of \$Q_1\$ to decline to oppose this change. Turns out that a simple voltage divider does this job well. All we need to do is to make sure that the current through the voltage divider is a lot more than the required base current of \$Q_2\$, so that when \$Q_2\$ adjusts its collector current and needs more (or less) base current, that this doesn't affect the divider voltage (much.)

That's really the essence of it. I added those two diodes to help stabilize things vs ambient temps. But they aren't strictly necessary if you don't mind your voltage rails shifting around a little more with temperature. As it is, they may still drift around by maybe \$\tfrac{25\:\text{mV}}{^{\circ}C}\$, just doing a short loop-around bit of guess-work. But if you don't mind it being twice as bad then you can replace the resistor and two diodes with a simple resistor, instead:



simulate this circuit
The actual value of \$R_3\$ may need to be adjusted a bit here, as we don't actually know just how much base current is needed (probably less than I guessed -- a lot less.) So perhaps closer to the \$12\:\text{k}\Omega\$ value? But you can use a potentiometer here, I suppose, to make this adjustable, too.