Solve $x^2+2=y^3$ using infinite descent?

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.

Even though you've already accepted my other answer (to your second/last question), I thought I’d add some thoughts on how the equation $x^2+2=y^3$ might be attacked by descent.

This answer is Method #1. I would love to brainstorm how this might be completed, or why it cannot.

Rewrite the equation as \begin{align} x^2+3 &= y^3+1 = (y+1)(y^2-y+1). \end{align} We can show that $3 \nmid x(y+1)$, and that $x,y$ are odd, and hence that $\gcd(y+1,y^2-y+1)=1$. Therefore by well-known results, we have \begin{align*} y+1 &= a^2 + 3b^2, \\ y^2-y+1 &= c^2 + 3d^2 \end{align*} for integers $a,b,c,d$ with $ac \ne 0$. (Since $y^2-y+1 = \tfrac{1}{4}\bigl((y+1)^2 +3((y+1)-2)^2\bigr)$, we can actually define $c,d$ in terms of $a,b$, but for now this should give you an adequate idea of my suggested approach.) Multiplying gives \begin{align} x^2 + 3 = (a^2+3b^2)(c^2+3d^2) = (ac \pm 3bd)^2 + 3(ad \mp bc)^2, \end{align} which can be rewritten as \begin{align} x^2 - 3(ad \mp bc)^2 = (ac \pm 3bd)^2 - 3(1)^2 = k, \tag{$\star$} \end{align} where $k \le x^2-3$ is an unknown integer. Now ($\star$) gives two solutions to the equation $U^2-3V^2 = k$, and $(ac \pm 3bd,1)$ with one of the signs is the fundamental solution (because of the $1$ at the end).

My intuition says that we can apply some sort of descent on ($\star$), and ultimate deduce that $a=\pm 2$ and $b=0$, forcing $y=a^2+3b^2-1=4+0-1=3$, as desired.

See this thread for more.

Here’s Potential Descent Mechanism #2 for discussion.

Clearly, $x$ and $y$ are both odd, and $x > y$. Hence there exist integers $a > b \ge 1$ such that $x=a+b$ and $y=a-b$. After substitution, you end up with a new third-degree equation, which I believe to be more susceptible to attack due to the number of terms and cross terms.

See this MSE thread, and this one, and this MO thread for examples of me trying — unsuccessfully — to apply higher-degree Vieta-jumping to obtain a descent path against such equations. I still think it's possible.