Why doesn't H₄O²⁺ exist?
I myself was always confused why $\ce{H3O^+}$ is so well-known and yet almost nobody talks of $\ce{H4O^2+}$. I mean, $\ce{H3O^+}$ still has a lone pair, right? Why can't another proton just latch onto that? Adding to the confusion, $\ce{H4O^2+}$ is very similar to $\ce{NH4+}$, which again is extremely well-known. Even further, the methanium cation $\ce{CH5+}$ exists (admittedly not something you'll find on a shelf), and that doesn't even have an available lone pair!

It is very useful to rephrase the question "why is $\ce{H4O^2+}$ so rare?" into "why won't $\ce{H3O^+}$ accept another proton?". Now we can think of this in terms of an acid-base reaction:

$$\ce{H3O^+ + H+ -&gt; H4O^2+}$$

Yes, that's right. In this reaction $\ce{H3O^+}$ is the base, and $\ce{H^+}$ is the acid. Because solvents can strongly influence the acidity of basicity of dissolved compounds, and because inclusion of solvent makes calculations tremendously more complicated, we will restrict ourselves to the gas phase (hence $\ce{(g)}$ next to all the formulas). This means we will be talking about proton affinities.

Before we get to business, though, let's start with something more familiar:

$$\ce{H2O(g) + H+(g) -&gt; H3O^+(g)}$$

Because this is in the gas phase, we can visualise the process very simply. We start with a lone water molecule in a perfect vacuum. Then, from a very large distance away, a lone proton begins its approach. We can calculate the potential energy of the whole system as a function of the distance between the oxygen atom and the distant proton. We get a graph that looks something like this:



For convenience, we can set the potential energy of the system at 0 when the distance is infinite. At very large distances, the lone proton only very slightly tugs the electrons of the $\ce{H2O}$ molecule, but they attract and the system is slightly stabilised. The attraction gets stronger as the lone proton approaches. However, there is also a repulsive interaction, between the lone proton and the nuclei of the other atoms in the $\ce{H2O}$ molecule. At large distances, the attraction is stronger than the repulsion, but this flips around if the distance is too short. The happy medium is where the extra proton is close enough to dive into the molecule's electron cloud, but not close enough to experience severe repulsions with the other nuclei.

In short, a lone proton from infinity is attracted to a water molecule, and the potential energy decreases up to a critical value, the bond length. The amount of energy lost is the proton affinity: in this scenario, a mole of water molecules reacting with a mole of protons would release approximately $\mathrm{697\ kJ\ mol^{-1}}$ (values from this table). This reaction is highly exothermic

Alright, now for the next step:

$$\ce{H3O^+(g) + H+(g) -&gt; H4O^2+(g)}$$

This should be similar, right? Actually, no. There is a very important difference between this reaction and the previous one; the reagents now both have a net positive charge. This means there is now a strong additional repulsive force between the two. In fact, the graph above changes completely. Starting from zero potential at infinity, instead of a slow decrease in potential energy, the lone proton has to climb uphill, fighting a net electrostatic repulsion. However, even more interestingly, if the proton does manage to get close enough, the electron cloud can abruptly envelop the additional proton and create a net attraction. The resulting graph now looks more like this:



Very interestingly, the bottom of the "pocket" on the left of the graph (the potential well) can have a higher potential energy than if the lone proton was infinitely far away. This means the reaction is endothermic, but with enough effort, an extra proton can be pushed into the molecule, and it gets trapped in the pocket. Indeed, according to Olah et al., J. Am. Chem. Soc. 1986,  108 (5), pp 1032-1035, the formation of $\ce{H4O^2+}$ in the gas phase was calculated to be endothermic by $\mathrm{248\ kJ\ mol^{-1}}$ (that is, the proton affinity of $\ce{H3O^+}$ is $\mathrm{-248\ kJ\ mol^{-1}}$), but once formed, it has a barrier towards decomposition (the activation energy towards release of a proton) of $\mathrm{184\ kJ\ mol^{-1}}$ (the potential well has a maximum depth of $\mathrm{184\ kJ\ mol^{-1}}$).

Due to the fact that $\ce{H4O^2+}$ was calculated to form a potential well, it can in principle exist. However, since it is the product of a highly endothermic reaction, unsurprisingly it is very hard to find. The reality in solution phase is more complicated, but its existence has been physically verified (if indirectly).

But why stop here? What about $\ce{H5O^3+}$?

$$\ce{H4O^2+(g) + H+(g) -&gt; H5O^3+(g)}$$

I've run a rough calculation myself using computational chemistry software, and here it seems we really do reach a wall. It appears that $\ce{H5O^3+}$ is an unbound system, which is to say that its potential energy curve has no pocket like the ones above. $\ce{H5O^3+}$ could only ever be made transiently, and it would immediately spit out at least one proton. The reason here really is the massive amount of electrical repulsion, combined with the fact that the electron cloud can't reach out to the distance necessary to accommodate another atom.

You can make your own potential energy graphs here. Note how depending on the combination of parameters, the potential well can lie at negative potential energies (an exothermic reaction) or positive potential energies (an endothermic reaction). Alternatively, the pocket may not exist at all - these are the unbound systems. 

EDIT: I've done some calculations of proton affinities/stabilities on several other simple molecules, for comparison. I do not claim the results to be quantitatively correct.



$$
\begin{array}{lllll}
\text{Species} &amp; \ce{CH4}  &amp; \ce{CH5+}    &amp; \ce{CH6^2+}     &amp; \ce{CH7^3+}     &amp; \ce{CH8^4+}     \\
    \text{Stable in gas phase?} &amp; \text{Yes} &amp;  \text{Yes}   &amp; \text{Yes} &amp; \text{Yes}  &amp; \text{No} \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  &amp; 556 &amp; -246 &amp; -1020   &amp; N/A   &amp; N/A  \\
\end{array}
$$

Notes:


Even without a lone pair, methane ($\ce{CH4}$) protonates very exothermically in the gas phase. This is a testament to the enormous reactivity of a bare proton, and the huge difference it makes to not have push a proton into an already positively-charged ion.
For most of the seemingly hypercoordinate species in these tables (more than four bonds), the excess hydrogen atoms "pair up" such that it can be viewed as a $\ce{H2}$ molecule binding sideways to the central atom. See the methanium link at the start.




$$
\begin{array}{lllll}
\text{Species} &amp; \ce{NH3}  &amp; \ce{NH4+}    &amp; \ce{NH5^2+}     &amp; \ce{NH6^3+}      \\
    \text{Stable in gas phase?} &amp; \text{Yes} &amp;  \text{Yes}   &amp; \text{Yes} &amp; \text{No}  \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  &amp; 896 &amp; -410 &amp; N/A   &amp; N/A    \\
\end{array}
$$

Notes:


Even though the first protonation is easier relative to $\ce{CH4}$, the second one is harder. This is likely because increasing the electronegativity of the central atom makes the electron cloud "stiffer", and less accommodating to all those extra protons.
The $\ce{NH5^{2+}}$ ion, unlike other ions listed here with more than four hydrogens, appears to be a true hypercoordinate species.  Del Bene et al. indicate a five-coordinate square pyramidal structure with delocalized nitrogen-hydrogen bonds.




$$
\begin{array}{lllll}
\text{Species} &amp; \ce{H2O}  &amp; \ce{H3O+}    &amp; \ce{H4O^2+}     &amp; \ce{H5O^3+}      \\
    \text{Stable in gas phase?} &amp; \text{Yes} &amp;  \text{Yes}   &amp; \text{Yes} &amp; \text{No}  \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  &amp; 722 &amp; -236 &amp; N/A   &amp; N/A    \\
\end{array}
$$

Notes:


The first series which does not accommodate proton hypercoordination.
$\ce{H3O+}$ is easier to protonate than $\ce{NH4+}$, even though oxygen is more electronegative. This is because the $\ce{H4O^2+}$ nicely accommodates all protons, while one of the protons in  $\ce{NH5^2+}$ has to fight for its space.




$$
\begin{array}{lllll}
\text{Species} &amp; \ce{HF}  &amp; \ce{H2F+}    &amp; \ce{H3F^2+}     &amp; \ce{H4F^3+}     \\
    \text{Stable in gas phase?} &amp; \text{Yes} &amp;  \text{Yes}   &amp; \text{Yes} &amp; \text{No}  \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  &amp; 501 &amp; -459 &amp; N/A   &amp; N/A    \\
\end{array}
$$

Notes:


Even though $\ce{H3F^2+}$ still formally has a lone pair, its electron cloud is now so stiff that it cannot reach out to another proton even at normal bonding distance.




$$
\begin{array}{lllll}
\text{Species} &amp; \ce{Ne}  &amp; \ce{NeH+}    &amp; \ce{NeH2^2+}    \\
    \text{Stable in gas phase?} &amp; \text{Yes} &amp;  \text{Yes}   &amp; \text{No}   \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  &amp; 204 &amp; N/A   &amp; N/A    \\
\end{array}
$$

Notes:


$\ce{Ne}$ is a notoriously unreactive noble gas, but it too will react exothermically with a bare proton in the gas phase.
Depending on the definition of electronegativity used, it is possible to determine an electronegativity for $\ce{Ne}$, which turns out to be even higher than $\ce{F}$. Accordingly, its electron cloud is even stiffer.




$$
\begin{array}{lllll}
\text{Species} &amp; \ce{H2S}  &amp; \ce{H3S+}    &amp; \ce{H4S^2+}     &amp; \ce{H5S^3+}     &amp; \ce{H6S^4+}     \\
    \text{Stable in gas phase?} &amp; \text{Yes} &amp;  \text{Yes}   &amp; \text{Yes} &amp; \text{Yes}  &amp; \text{No} \\
\text{Approximate proton affinity}\ (\mathrm{kJ\ mol^{-1}})  &amp; 752 &amp; -121 &amp; -1080   &amp; N/A   &amp; N/A  \\
\end{array}
$$

Notes:


The lower electronegativity and larger size of $\ce{S}$ means its electrons can reach out further and accommodate protons at a larger distance, while reducing repulsions between the nuclei. Thus, in the gas phase, $\ce{H2S}$ is a stronger base than $\ce{H2O}$. The situation is inverted in aqueous solution due to uniquely strong intermolecular interactions (hydrogen bonding) which are much more important for $\ce{H2O}$.
$\ce{H3S+}$ also has an endothermic proton affinity, but it is lower than for $\ce{H3O+}$, and therefore $\ce{H4S^2+}$ is easier to make. Accordingly, $\ce{H4S^2+}$ has been detected in milder (though still superacidic!) conditions than $\ce{H4O^2+}$.
The larger size and lower electronegativity of $\ce{S}$ once again are shown to be important; the hypercoodinate $\ce{H5S^3+}$ appears to exist, while the oxygen analogue doesn't.