This problem arose when considering storage of cannonballs in ndimensional pirate ships, as explained in this MSE post. This MO question 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 $a_1\in\{\lfloor\frac{n}{\ln(2)}\rfloor,\lfloor\frac{n}{\ln(2)}\rfloor1\}$, 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.

$\begingroup$ For $n = 3$ this should follow from the fact that the MordellWeil group of the curve $x^3  2y^3 = 1$ over $\mathbb{Q}$ is of rank $0$. $\endgroup$ – WhatsUp Dec 13 '16 at 14:48

$\begingroup$ For $n = 4$ it is also easy to see that there is no nontrivial solution, by writing the equation as $(x^2 + 1)(x^21)=2y^4$. $\endgroup$ – WhatsUp Dec 13 '16 at 14:55
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/BCrelle2.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