This problem arose when considering storage of cannonballs in n-dimensional pirate ships, as explained in [this MSE post](https://math.stackexchange.com/questions/2042839/solutions-to-the-diophantine-equation-xn-2yn-1-can-the-sum-of-the-first-n). [This MO question](https://mathoverflow.net/questions/39561/is-there-an-elementary-way-to-find-the-integer-solutions-to-x2-y3-1) can also be reduced to the $n=3$ case. If $x,y$ is a solution then $$0<\frac{x}{y}-2^\frac1n<\frac{2^\frac1n}{2ny^n}$$ then by Roth's theorem this has finitely many solutions for fixed $n$. Let $$2^{1/n}=a\_0+\frac{1}{a\_1+\dots}$$
be the canonical continued fraction of $2^{1/n}$, then $a\_0=1$ and $a\_1\in\{\lfloor\frac{n}{\ln(2)}\rfloor,\lfloor\frac{n}{\ln(2)}\rfloor-1\}$, and since $\frac{x}{y}$ is a convergent of this continued fraction, $y>\frac{n}{\ln(2)}-1$. There are no solutions with $x^{n}<2^{64}$. It is also sufficient to only consider $n=4$ and odd primes, in FLT fashion.



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.


