The content is miles outside what I know about. So the question is a mixture of idle curiosity and maybe having this answered somewhere on the Internet. It is likely I will not be able to understand the answer.
How exactly does Wiles' proof of Fermat's Last Theorem fail for $n=2$?
tl;dr version: A solution $a^\ell + b^\ell = c^\ell$ would generate a weird (i.e,. not modular) elliptic curve over $\mathbb{Q}$ for prime $\ell \geq 5$. Wiles proved that sufficiently weird elliptic curves don't exist, so that means Fermat's Last Theorem holds for $\ell \geq 5$.
There isn't an easy or concise answer to the question, simply because the number theory involved is so complicated. It had been known for a while before Wiles that Taniyama-Shimura conjecture would imply Fermat's Last Theorem, and Wiles proved it for a large enough class of curves to also prove that theorem. (The conjecture is a much deeper result than Fermat's Last Theorem, and Wiles' proof was also extended to the general case a bit later.)
The idea behind the conjecture is that for an elliptic curve $E/\mathbb{Q}$, you can reduce mod $p$ for each prime $p$ and count the number of points on $E$ over $\mathbb{F}_p$. Explicitly, given a Weierstass form $E:y^2 = x^3 + a x + b$, you can just count number of points $(x, y)\pmod{p}$ with $y^2 \equiv x^3 + ax + b\pmod{p}$. These can be tossed into something like an Euler product or a Dirichlet series to get to a resulting object called an $L$-series. (Sort of, anyway. There are corrections that have to be made for a finite number of primes $p$). If the Taniyama-Shimura conjecture holds, then we can do a bit of manipulation to get a nonzero function $f$ that is extraordinarily symmetric under the the group \begin{align*} \Gamma_0(N) = \left\{\begin{pmatrix} a & b \\ c & d\end{pmatrix}\in SL_2(\mathbb{Z}):\, c\equiv 0\!\!\pmod{N}\right\} \end{align*} acting on $\{z\in \mathbb{C}:\, \operatorname{Im} z > 0\}$ by $\begin{pmatrix} a & b \\ c & d\end{pmatrix}.z = (az + b)/(cz + d)$, where $N$ is a certain integer attached to $E$. (Specifically, $f$ is a modular form, and so the Taniyama-Shimura conjecture is also called the modularity theorem. Also, what I'm calling $N$ here isn't quite the conductor of $E$, but it divides it.)
This condition on the action $\Gamma_0(N)$ is pretty strict, and there aren't many such $f$. In particular, for $N = 2$, there isn't any nonzero function $f$ at all. But it turns out that given coprime (and nonzero) $a, b, c$ with $a^\ell + b^\ell = c^\ell$ for prime $\ell \geq 5$, the curve $E:y^2 = x(x - a^\ell)(x - c^\ell)$ happens to have $N = 2$. That contradiction implies that no such $a, b, c$ exist; that is, Fermat's Last Theorem holds for $\ell$. It's then clear that Fermat's Last Theorem holds for any exponent $n$ divisible by some prime $\ell \not = 2, 3$. But the cases $n = 3, 4$ are classical, and so the theorem holds for all $n > 2$.
So, what happens with $\ell = 2$? Well, the theorem obviously fails: There are plenty of points with $a^2 + b^2 = c^2$, and we're going to have to get a different value of $N$ for $\ell = 2$ in order for Taniyama-Shimura to still hold. There isn't an easy way to describe this part of the proof, but the basic idea is to compute the conductor of a certain elliptic curve, and there are a couple of cases describing its behavior when reduced mod $2$. If $\ell \geq 5$, we can eliminate all but a few easy cases and get an explicit value for $N$. That's probably not particularly useful, but see the caveat above that there isn't an easy or concise answer to this question. The major requirement of the proof is that $\ell$ be sufficiently large; the fact that $\ell \geq 5$ works is a happy coincidence.
If you're looking for a deeper reason why $\ell = 2$ fails, then the reason is that the Fermat curves $C:X^\ell + Y^\ell - Z^\ell$ are very different for $\ell = 2$ than for (prime) $\ell > 2$. The genus of $C$ is $g = \frac{1}{2}(\ell - 1)(\ell - 2) $, which is greater than $1$ for $\ell > 3$. Falting's Theorem states that such curves with $g > 1$ only have finitely many points where you can take the coordinates to all be rational. (The case $\ell = 3$ is different but easily handled classically.) For $\ell = 2$, this is clearly not the case; there are infinitely many Pythagorean triples. Now, "finitely many" is unfortunately a very long way from "zero," but that's the fundamental reason why $\ell = 2$ is different. The reason why Wiles' proof fails is that the Frey curve $E:y^2 = x(x - a^\ell)(x - c^\ell)$ gives you a different value of $N$ when $\ell = 2$, and we need $N = 2$ to get a violation of Taniyama-Shimura. As for why that value is different...well, it's a function of $\ell$, and it just happens to be different for $\ell = 2$ versus $\ell \geq 5$. I'm not sure there's anything really deep going on there.
I'm skipping over and simplifying a lot of the details here, but there's a more thorough but readable treatment in, e.g., the last chapter of Knapp's "Elliptic Curves."
The other "accepted" answer in my opinion more or less misses the point. So here is another answer.
Consider the elliptic curve
$$E: y^2 = x(x - a^p)(x + b^p).$$
The $p$-torsion points $E[p]$ of $E$ form a $2$-dimensional vector space over $\mathbf{F}_p$ which has an action of the Galois group, and this gives rise to a representation
$$\overline{\rho}: G_{\mathbf{Q}} \rightarrow \mathrm{GL}_2(\mathbf{F}_p).$$
The structure of the proof is based on the idea that the shape of $E$ implies that $\overline{\rho}$ has very special ramification properties. Assuming that $E$ and hence $\overline{\rho}$ is modular, Ribet's argument shows that either $\overline{\rho}$ should come from another modular form of a weight and level which doesn not exist (which is a contradiction, and this is the part which uses modularity) or $\overline{\rho}$ is not absolutely irreducible.
In particular, in order to complete the proof, one has to rule out the possibility that $E[p]$ is not absolutely irreducible. Assuming one can massage the choices of $a$ and $b$ to guarantee that $E$ is semistable, this turns out to be equivalent to showing that $E$ does not have a ratio nal point of order $p$. But since $E$ already has rational $2$-torsion, this is not possible for primes $p \ge 5$ by Mazur's theorem (https://en.wikipedia.org/wiki/Torsion_conjecture). Without Mazur's theorem (or something equivalent or stronger) you can't prove Fermat.
Returning to the case $p = 2$, the key reason the argument fails is that $\overline{\rho}$ can be (and indeed is) reducible, but knowing that $E[2]$ is reducible is not any contradiction to Mazur's theorem. In particular, you can write down $E: y^2 = x(x-9)(x+16)$ and then this is a perfectly good elliptic curve with rational $2$-torsion points.
$(1)$ We assume that Fermat's Last Theorem is false i.e. there exists some $a,b,c > 0$ and prime $p > 2$ s.t. $a^p + b^p = c^p$.
$(2)$ It follows from Ribet's Theorem that the elliptic curve $y^2 = x(x-a^p)(x + b^p)$ (known as the Frey Curve) has no associated modular form.
$(3)$ However, it follows from the Modularity Theorem that every rational elliptic curve is modular.
$(4)$ This is a contradiction.
The reason that Fermat's Last Theorem doesn't hold for $p=2$ is that $(2)$ doesn't hold for $p = 2$.
– Jun 03 '22 at 11:08