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What is it, in Mihailescu's proof of Catalan conjecture, that uses explicitly the fact that there is a 1 on the right hand side of $x^p - y^q = 1$? In other words, why can't we extend his argument to prove stuff about, say, $x^p - y^q = 2$?

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    $\begingroup$ The book Schoof: "Catalan's conjecture", written in 2007, says on page 8 that it is a research problem to solve equation $x^p-y^q=m$ for $m=2,6,10,14,18,\ldots$. $\endgroup$ Commented May 23, 2014 at 14:44

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This article by Tauno Metsänkylä gives a good explanation of Mihăilescu's proof. There are some crucial steps in the proof that could not be done in the case $x^p-y^q=2,$ say. The idea is to write $$\frac{x^p-1}{x-1}(x-1)=y^q$$ and observe that if $x-1$ and $\frac{x^p-1}{x-1}$ are coprime, we arrive at $x-1=a^q, \quad \frac{x^p-1}{x-1}=b^q$ for some integers $a$ and $b$ (this could not be done with the equation $x^p-2=y^q$ as $x-2$ does not usually divide $x^p-2$). This is the so called first case that was solved by J. W. S. Cassels in 1960. Hence Mihăilescu was left with the second case where $\gcd(\frac{x^p-1}{x-1},x-1)=p$, so $$\frac{x^p-1}{x-1}=pb^q.$$ Denoting by $\zeta$ a primitive $p$th root of unity, we get $$\prod_{k=1}^{p-1}\frac{x-\zeta^k}{1-\zeta^k}=b^q$$ (again $\frac{x^p-2}{(x-2)p}$ does not have such a nice factorization). By moving to ideals, we get the equation $$\left\langle\frac{x-\zeta^k}{1-\zeta^k}\right\rangle=J^q,$$ where $J$ is an ideal of $\mathbb{Z}[\zeta]$. One of Mihăilescu's key ideas was to transfer this equation back to an equation of numbers by using an annihilator $\theta$ of $\mathbb{Z}[\zeta]$ to arrive at $$\left(\frac{x-\zeta^k}{1-\zeta^k}\right)^{\theta}=\varepsilon \gamma^q,$$ where $\varepsilon$ is a unit in $\mathbb{Z}[\zeta]$ and $\gamma\in \mathbb{Q}(\zeta)$. From this equation, he was able to derive his miraculous congruences $$x\equiv 0 \pmod{q^2},\quad p^{q-1}\equiv 1 \pmod{q^2}, \quad y\equiv 0 \pmod{p^2},\quad q^{p-1}\equiv 1 \pmod{p^2},$$ which are a key to the proof. The equation $x^p-y^q=2$ may actually be an open problem; at least I could not find any references on it. But of course if we accept Mochizuki's proof of the abc-conjecture, it has only finitely many solutions.

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