Basic BJT current mirror : ... so the collector-base voltage of Q2 is also zero?
Without getting too involved in the details, there are typically two very important and one slightly less important (usually) details to keep track of with respect to a simplified BJT model. (Ignoring temperature differences.) These are the saturation current, \$I_\text{SAT}\$, which associates the relationship between \$V_\text{BE}\$ and \$I_\text{C}\$; the value of \$\beta\$, which associates the relationship between \$I_\text{B}\$ and \$I_\text{C}\$; and a parameter describing the Early effect, \$V_A\$, which modifies the earlier relationship between \$V_\text{BE}\$ and \$I_\text{C}\$ by including an effect due to \$V_\text{CE}\$.
If you can hold just the first two ideas in mind at one time, you can analyze many circuits close enough for usable results. The third idea, the Early effect, can become important in other circuits. This one is an example where it often may matter.

This is a good point to suggest that you also look at the
cascode arrangement where one BJT is used in a way to pass along the collector current of another BJT while also mitigating the Early
effect in that other BJT. But I won't discuss it here.

So let's look at the schematic, with some added bits to help clarify as I write more:



simulate this circuit &ndash; Schematic created using CircuitLab
On the left, I've added two resistors. These are due to the Early effect. Their values can be approximated with \$r_o\approx \frac{V_A}{I_\text{C}}\$. On the right, I've simplified the diagram, slightly. Since \$Q_1\$ is diode-connected, we can replace it with a diode symbol. This may help get some points across more easily. (Or it may do the opposite. We'll see.)
The first part of the quote from your link is:

If \$Q_1\$ and \$Q_2\$ are matched,

For the most part, by matched they mean the first parameter I mentioned: \$I_\text{SAT}\$. This is the most important parameter to match because of the Shockley equation relating \$V_\text{BE}\$ and \$I_\text{C}\$ (lacking the Early effect):
$$I_\text{C}=I_\text{SAT}\cdot\left(e^{^{\left[\frac{V_\text{BE}}{V_T}\right]}}-1\right)$$
You can solve the above for \$V_\text{BE}\$, as well:
$$V_\text{BE}=V_T\cdot \ln\left(\frac{I_\text{C}}{I_\text{SAT}}+1\right)$$
If you know one, you can calculate the other.
The current mirror depends upon this fact. By letting \$R\$ set \$Q_1\$'s \$I_\text{C}\$, it responds by generating a \$V_\text{BE}\$. It can do this, because the base is tied to the collector and will draw the \$I_\text{B}\$ needed to handle base recombination for \$I_\text{C}\$. (If you assume that \$\beta\to\infty\$, then you can temporarily ignore \$I_\text{B}\$ and therefore not worry about \$\beta\$.)
Since \$Q_1\$'s generated \$V_\text{BE}\$ is passed along to \$Q_2\$'s \$V_\text{BE}\$, this implies that \$Q_2\$'s \$I_\text{C}\$ will be the same.
All this assumes that both BJTs are operating in their active regions, of course.

\$I_\text{SAT}\$ is the most important parameter to match because of
the current mirror's ratio of currents is
\$\frac{I_{\text{C}_2}}{I_{\text{C}_1}}\approx
 \frac{I_{\text{SAT}_2}}{I_{\text{SAT}_1}}\$. (You can derive that
relatively easily from the above equations and using the given current
mirror arrangement above.)
Discrete BJTs from the same device family can have their values of
\$I_\text{SAT}\$ vary by as much as a factor of 3 around the average
value. This translates to as much as \$\pm 30\:\text{mV}\$ difference in \$V_\text{BE}\$ given the
same \$I_\text{C}\$. If you selected poorly enough, getting one BJT
from one end of the spectrum and another one from the opposite end,
the difference in current could be as much as 8 or 9. Not so good for
a current mirror where you want very similar currents. A way to help
repair this is to add emitter resistors in discrete designs. But how
to design that so it works well is beyond the scope here.
Manufacturers offer BJT pairs. Some are just offered in pairs for
convenience and there's no matching intended beyond what accidentally
occurs. Some are offered with varying degrees of
\$V_\text{BE}\$-matching. Some are offered with very good degrees of
\$V_\text{BE}\$-matching and then also reasonably good degrees of
\$\beta\$-matching, as well. (The latter costs more, of course.)

The next part of the quote from your link is:

that is, have substantially the same device properties,

I've already discussed the important property, \$I_\text{SAT}\$. But I've left out \$\beta\$, above. So I'll address it quickly.
\$\beta\$ is a device property and not all of the current supplied by \$R\$ makes it to the collector of \$Q_1\$. If the value of \$\beta\$ is different between the two BJTs, or if it is the same between them, the fact is that this causes an error in the estimate of \$Q_2\$'s \$I_\text{C}\$, since \$R\$ has to supply two base currents as well as \$Q_1\$'s \$I_\text{C}\$. (This can be greatly aided by the addition of another BJT, not shown.)
Assuming the BJTs are matched on \$I_\text{SAT}\$ and if the \$\beta\$ of both BJTs are exactly the same, the current in \$R\$ still must be reduced by \$2\cdot \frac{I_\text{C}}{\beta}\$ in order to figure out what remains for \$Q_1\$'s \$I_\text{C}\$. Again assuming the BJTs are matched on \$I_\text{SAT}\$ and if the \$\beta\$ of the BJTs are different, then the current in \$R\$ must be reduced by \$\frac{I_{\text{C}}}{\beta_1}+\frac{I_{\text{C}}}{\beta_2}\$ in order to figure out what remains for \$Q_1\$'s \$I_\text{C}\$. (Both will still have the same \$I_\text{C}\$ if they are matched on \$I_\text{SAT}\$.) This is just a matter of figuring out \$Q_1\$'s \$I_\text{C}\$ and the values of \$\beta\$ are typically high enough (and can be repaired by the addition of another BJT, anyway), so I don't really count it as an issue here. But do be aware of it, anyway.
The next part of the quote from your link is:

and if the mirror output voltage is chosen so
the collector-base voltage of \$Q_2\$ is also zero,

The point here is addressing itself now to the Early effect. (Those resistors I show in the schematic.) \$Q_1\$ is being operated with its base voltage exactly equal to its collector voltage. So this means, by definition, that \$V_\text{BC}=0\:\text{V}\$ for \$Q_1\$. So \$V_\text{CE}=V_\text{BE}\$, also. Since \$V_{\text{CE}_1}\$ is impressed across \$r_{o_{_1}}\$, it follows that there is a small current through it and this adds to what is observed as \$Q_1\$'s \$I_\text{C}\$. The same thing happens with \$Q_2\$. So if you want to observe the same \$I_\text{C}\$ in \$Q_2\$, then \$Q_2\$'s \$V_\text{CE}\$ must also be the same. But this means that \$Q_2\$'s  \$V_\text{BC}=0\:\text{V}\$, as well. That's why they wrote that.
In the above schematics, this would mean that \$V_\text{X}= V_\text{BE}\$. In reality, this almost never happens (unless you use one of variations of a Wilson mirror.) So if, say, you set \$V_\text{X}=V_\text{CC}\$ and you place a resistor in between \$Q_2\$'s collector and \$V_\text{X}\$, then that resistor will drop some voltage due to the current through it. That voltage may, or may not, set the collector voltage of \$Q_2\$ to be exactly the same as \$Q_1\$'s collector voltage. If not, then \$r_{o_{_2}}\$ will sink more current than \$Q_1\$'s \$r_{o_{_1}}\$ does. And this will be yet another unaccounted error.
So they are just being cautious in their wording.

In the interest of some completeness, I'll present some common modifications. You should become familiar with them:



simulate this circuit
In the left-most case above, I've added \$Q_3\$. \$Q_3\$ reduces the error current stolen from \$R_\text{SET}\$, used to supply the two base currents for \$Q_1\$ and \$Q_2\$. That's because \$Q_3\$'s emitter current is doing that job and only a fraction of it (divided by \$Q_3\$'s \$\beta+1\$) is taken from the current supplied by \$R_\text{SET}\$. That will be a lot less than before. The price to pay here is that now \$Q_1\$'s collector voltage has to be \$V_\text{BE}\$ higher. This is easily accounted for when determining the value of \$R_\text{SET}\$, so it's not considered a serious problem but rather a useful improvement.
In the middle case, \$Q_3\$ is moved over into a new position, pushed over so that it falls within the load current path, now. But positioned here, it serves useful two jobs at once. It still supplies the two base currents for \$Q_1\$ and \$Q_2\$. But it also arranges things so that \$Q_2\$'s collector voltage is the same as its base voltage. This effectively nullifies \$Q_2\$'s Early effect. But now \$Q_1\$'s collector voltage isn't the same as its base voltage, but instead one \$V_\text{BE}\$ above its base voltage. So \$Q_1\$ and \$Q_2\$ are operating with slightly different \$V_\text{CE}\$ values and the Early effect has a small effect on the collector current differences. But as the difference isn't that much the problem is relatively small. Another impact is that now \$R_\text{SET}\$ is again supplying more substantial base current to \$Q_3\$. But this is still better than in the case with only \$Q_1\$ and \$Q_2\$, where \$R_\text{SET}\$ had to supply two base currents instead of just one, as in this new case. This is the basic Wilson mirror.
The right side shows what I call a full Wilson mirror. If you look closely, you'll see now that the collectors of both \$Q_1\$ and \$Q_2\$ are identical. (This is very good, as the Early effect will be nearly identical for them.) \$R_\text{SET}\$ now has to again supply two base currents. But as that can be accounted for when setting a value for it, the benefits are considered worth the costs in some circumstances. You could, if you wanted to (and had the extra voltage headroom for it and liked slapping down more BJTs), place a new \$Q_5\$ in a similar position to \$Q_3\$ shown in the left-most schematic to alleviate the base currents drawn away from \$R_\text{SET}\$. But I don't recall seeing it used.
Please note that \$Q_3\$ in the middle and right schematics is operated as a cascode BJT. It passes along \$Q_2\$'s collector current, which is no longer impacted by collector voltage variations as it is isolated from them by \$Q_3\$. You might think that \$Q_3\$ should still suffer from the Early effect, as it's collector voltage can vary depending on the load and the current setting. But since it doesn't control its collector current, as this is set by \$Q_2\$'s collector current, it doesn't matter anymore. It simply passes long the current to the load, regardless.
There are some remaining nuances which would disagree (but only slightly) with what I presented above. But it would draw this out way beyond where I should. Feel free to think about it on your own.