Are there any solutions to the diophantine equation $x^n-2y^n=1$ with $x>1$ and $n>2$?

Delone (1930) and Nagell (1928) showed for any nonzero integer $d$ that the equation $x^3 - dy^3 = 1$ has at most one solution in integers $(x,y)$ besides $(1,0)$, with no constraint on the signs of $x$ and $y$. In particular, since $x^3 - 2y^3 = 1$ has the integral solution $(-1,-1)$, there is no integral solution $(x,y)$ in positive integers.

This theorem was extended to exponent 4 by Ljunggren (1942) and to exponent 5 and higher by Bennett (2001): for $n \geq 3$ and $d \not= 0$, the equation $|x^n - dy^n| = 1$ has at most one solution in positive integers. See Theorem 1.1 of https://www.math.ubc.ca/~bennett/B-Crelle2.pdf (which actually treats a slightly more general equation). In particular, $|x^n - 2y^n| = 1$ has at most one solution $(x,y)$ in positive integers. Since $(x,y) = (1,1)$ fits, it is the only one. Of course $x^n - 2y^n = -1$ when $(x,y) = (1,1)$, so for $n \geq 3$ there is no solution to $x^n - 2y^n = 1$ when $x$ and $y$ are positive integers.

The case $n = 4$ and so $n$ even can be done by hand. Darmon and Merel proved (in 1997) the stronger statement that there aren't even any rational solutions to this equation for $n$ odd besides $(x^n,y^n) = (1,1)$, See their paper "Winding quotients and some variants of Fermat's last theorem," which can be found here:

http://www.math.mcgill.ca/darmon/pub/Articles/Research/18.Merel/paper.pdf