Using the fact that $\mathbb{Z}[\varphi]$ is a unique factorization domain in which we can decompose
$$
x^5+y^5=(x+y)(x^2+y^2-\varphi xy)(x^2+y^2-\bar\varphi xy),\quad\qquad{\rm(1)}
$$
we can give a proof of Fermat's Last Theorem for the case of exponent 5. Here, I am using a bar over a number to denote conjugation, so $\varphi=(1+\sqrt{5})/2$ and $\bar\varphi=(1-\sqrt{5})/2$.
Theorem: There are no solutions to $x^5+y^5=z^5$ for nonzero $x,y,z$ in $\mathbb{Z}[\varphi]$.
That is, for exponent 5, FLT holds in the ring $\mathbb{Z}[\varphi]$ and, in particular, it holds in the integers.
Before going any further, let's note a few facts about factorization in $\mathbb{Z}[\varphi]$. As is well known, it is norm-Euclidean, so is a unique factorization domain. We have the prime factorizations $5=(\sqrt{5})^2$ and $11=q\bar q$, where I am setting $q=4-\sqrt{5}$ (for the remainder of this post). The identity $\varphi\bar\varphi=-1$ shows that $\varphi$ is a unit. In fact, it is a fundamental unit, so that every unit in $\mathbb{Z}[\varphi]$ is of the form $\pm\varphi^r$ for integer $r$. It will also be useful to use mod-q arithmetic (with $q$ as above). Then, $\varphi=(1+\sqrt{5})/2=8$ (mod q). Therefore every element of the quotent $\mathbb{Z}[\varphi]/(q)$ is equal to a rational integer mod q. As 11 = 0 mod q, this gives $\mathbb{Z}[\varphi]/(q)\cong\mathbb{Z}/(11)$. So, mod-q arithmetic in $\mathbb{Z}[\varphi]$ is exactly the same as mod-11 arithmetic in the integers. In particular, every 5'th power is equal to one of $0,1,-1$ mod q. Applying this to the equation $x^5+y^5=z^5$ shows that at least one of $x,y,z$ must have a factor of q. By dividing through by their highest common factor, we reduce to the case where $x,y,z$ are coprime, so exactly one is a multiple of q. Rearranging as $(-z)^5+y^5=(-x)^5$ if necessary, we can always bring the multiple of q to the right hand side. This reduces the problem to the following.
Theorem 2: There are no solutions to $x^5+y^5=uz^5$ for nonzero coprime $x,y,z\in\mathbb{Z}[\varphi]$ with $u\in\mathbb{Z}[\varphi]$ a unit and $q$ dividing $z$.
Let's prove this by showing that, if we have one solution, then we can find another solution for which $xyz$ has strictly fewer distinct prime factors. Applied to a minimal solution, this would give a contradiction. This is essentially the method of descent used by Fermat himself for the case of exponent 4.
So, suppose we have one solution. Writing $c_0=x+y$, $c_1=x^2+y^2-\varphi xy$ and $c_2=x^2+y^2-\bar\varphi xy$, (1) gives the decomposition $uz^5=c_0c_1c_2$. Also,
$$
c_0^2-\bar\varphi c_1-\varphi c_2=0.\qquad\qquad{\rm(2)}
$$
We would like to show that the factors $c_0,c_1,c_2$ are 5'th powers, which will be easier if they are coprime. Using the fact that $x,y$ are coprime to $z$, the identities
$$
\begin{align}
&c_0^2-c_1=\sqrt{5}\varphi xy,\\
&c_0^2-c_2=-\sqrt{5}\bar\varphi xy,\\
&c_1-c_2=-\sqrt{5}xy
\end{align}
$$
show that the highest common factor of $c_0^2,c_1,c_2$ is either 1 or $\sqrt{5}$. Consider the case where $\sqrt{5}$ divides $z$. Then it will also divide at least one of $c_i$, and the identities above show that it divides each $c_i$. In particular, 5 divides $c_0^2$, so the identities above show that $\sqrt{5}$ divides each of $c_1,c_2$ exactly once.
In the case where $z$ is not a multiple of $\sqrt{5}$, let us set $\tilde c_0=c_0^2,\tilde c_1=c_1,\tilde c_2=c_2$ and, in the case where $\sqrt{5}$ divides $z$, set $\tilde c_0=c_0^2/\sqrt{5},\tilde c_1=c_1/\sqrt{5},\tilde c_2=c_2/\sqrt{5}$. These are coprime and
$$
\tilde c_0\tilde c_1^2\tilde c_2^2 = u^2\left(z^{2}/\sqrt{5}^{m}\right)^5
$$
where $m=0$ if $\sqrt{5}$ does not divide $z$ and $m=1$ if it does. As they are coprime, each prime factor of $z$ divides exactly one of the $\tilde c_i$, and its exponent is a multiple of 5. So, considering prime factorizations, each $\tilde c_i$ is equal to a unit multiplied by a fifth power $w_i^5$. So, (2) gives
$$
u_0w_0^5+u_1w_1^5+u_2w_2^5=0
$$
for units $u_i$. Without loss of generality, we assume that $q$ divides $w_0$ and, dividing through by $-u_1$ if necessary, we suppose that $u_1=-1$. Then, $u_2=\pm1$ mod q. However, being a unit, we have $u_2=\pm(\varphi)^r=\pm 8^r$ (mod q), and, looking at this mod 11, 5 must divide $r$. So, $u_2$ is a fifth power and, by absorbing $-u_2^{1/5}$ into $w_2$, we can take $u_2=-1$. So we have arrived at
$$
w_1^5+w_2^5=u_0w_0^5.
$$
Also, all prime factors of $w_0w_1w_2$ are factors of $z$. So, except in the case where $x,y$ are units, we have a solution with strictly fewer prime factors, and we are done.
So suppose that we have a solution to Theorem 2. Iteratively applying the procedure above will keep generating new solutions and, as the number of prime factors of $xyz$ cannot decrease indefinitely, we must eventually settle on the case where $x,y$ are units, so that $x/y=\pm\varphi^r$. Exchanging $x,y$ if necessary, we suppose that $r > 0$. Then, $q^5$ is a factor of $1\pm\varphi^{5r}$. Using the identity $\varphi^5=-1+\varphi^4q$, and applying the binomial identity, it can be seen that $rq$ must be a multiple of $q^{5}$, so $r$ is a multiple of $11^4$. In particular, $\vert x/y\vert=\vert\varphi\vert^r$ will be very large (note, $\vert\varphi\vert^{11^4} > 10^{3000}$). Then, the definitions above for $c_0^2,c_1,c_2$ are dominated by the $x^2$ terms, so the ratios $\tilde c_i/\tilde c_j$ are close to one. Going through these details bounds the ratios $w_i/w_j$ and, in particular, none of them will be as large as $\vert\varphi\vert^{11^4}$. This means that we cannot have $x,y$ and $w_1,w_2$ all units. So, continuing the induction will generate solutions with ever fewer prime factors, giving the required contradiction.
This method of approaching FLT for exponent 5 was something I came up with after seeing the exponent 3 case in lectures years ago. It is a bit tedious having to separately deal with the case where $x,y$ are units. Maybe that can be tidied up. Essentially, the reason why this method works is because $\mathbb{Z}[\varphi]$ consists precisely of the real algebraic integers of the cyclotomic field $\mathbb{Q}(\zeta_5)$.
Best Answer
To answer your last question: Yes, it could really be that only Fermat knew his method of descent [using contemporary techniques] well enough to make this problem submit to it.
The companion problem regarding $x^2+4=y^3$ also has no known descent proof, though he claimed to have one. There is no known descent proof of the fact that Pell's equation has infinite solutions — but Fermat claimed to have proven that by descent as well. In fact, of the ten problems mentioned in his letter to Carcavi, which Fermat claimed to prove by infinite descent, as far as I know only one (FLT for $n=3$) has had a published descent proof.
To summarize: If Fermat had only claimed to have proven one of his theorems (e.g. FLT) by descent, and no such proof was ever found, I would have no problem convincing myself that he was mistaken. But he claimed descent proofs of dozens of theorems, all of which were later proven true using other methods — at some point, we have to ask ourselves what he knew that we don't.