This problem arose when considering storage of cannonballs in n-dimensional 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 $a_1\in\{\lfloor\frac{n}{\ln(2)}\rfloor,\lfloor\frac{n}{\ln(2)}\rfloor-1\}$, 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.

This problem arose when considering storage of cannonballs in n-dimensional 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 $a_1\in\{\lfloor\frac{n}{\ln(2)}\rfloor,\lfloor\frac{n}{\ln(2)}\rfloor-1\}$, 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.

This problem arose when considering storage of cannonballs in n-dimensional 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 $a_1\in\{\lfloor\frac{n}{\ln(2)}\rfloor,\lfloor\frac{n}{\ln(2)}\rfloor-1\}$, 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.

This problem arose when considering storage of cannonballs in n-dimensional 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 $a_1\in\{\lfloor\frac{n}{\ln(2)}\rfloor,\lfloor\frac{n}{\ln(2)}\rfloor-1\}$, 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.