This is an attempt to generalize this question. $$x^n-2y^n=1\implies \frac{x}{y}=\left(2+\frac{1}{y^n}\right)^{\frac 1 n}=2^\frac1n\left(1+\frac{1}{2y^n}\right)^\frac1n<2^\frac 1n\left(1+\frac{1}{2ny^n}\right)$$ $$0<\frac{x}{y}-2^\frac1n<\frac{2^\frac1n}{2ny^n}$$ but since the irrationality measure of an algebraic number is $2$, this can only have finitely many solutions for a fixed $n>2$. Furthermore, any solutions would generate a suspiciously good rational approximation of $2^\frac1n$.
The argument used in the $n=3$ case can be easily generalized to prove that if none of $$x^n-2y^n=1$$ $$x^n-4y^n=1$$ $$x^n-6y^n=1$$ Have nontrivial solutions for $n>2$, then the sum of the first $n$ squares is never a perfect power other than $1$ and $4900$.
