number theory - Can we prove that there are no two perfect powers with difference $6$?

Here :

Are those lists known to be complete?

I asked whether the list of numbers given in the link is known to be complete. Since the generalized Catalan-conjecture is apparantly open, I think they are not. In particular I am interested in the smallest candidate, for which there might be no solution, the case $n=6$

Can we proof that there are no two perfect powers with difference $6$ ? If not, can we at least prove that such powers must be very large ?

It is clear that two perfect powers with difference $6$ have the same parity. Furthermore, the powers cannot be even because then both powers would be divisble by $4$, therefore the difference would be divisble by $4$ as well, which is impossible because $4$ does not divide $6$.

We also can easily see that $6$ cannot be the difference of two squares. And apparantly, the Mordell-curves $y^2=x^3+6$ and $y^2=x^3-6$ both have no integral solutions either.

Answer

There is no integer $c$ with absolute value exceeding $1$ for which we can even prove that the number of solutions to the equation
$$x^n-y^m=c$$

is finite (though such a result would follow trivially from the ABC-conjecture). The case with $c=6$ is indeed the smallest positive value where we expect there to be no solutions (which follows from a suitably explicit version of ABC), but, other than being able to show that there are no solutions to $x^2-y^n = \pm 6$ (and, presumably, being able to handle a few ``small'' pairs $m, n$), I don't believe there is anything more that we can prove with current technology.

The equations $x^2+6 = y^n$ and $x^2-6=y^n$ were solved by J. H. E.Cohn and by C. F. Barros (a student of Samir Siksek), respectively. The first of these is a relatively elementary argument (the paper is in Acta Arithmetic, from 1993) while the second is not.