Solve $a^n - b^n = 8$ with $a,b \in \mathbb{Z}$ and $n \in \mathbb{N}_{\geq 2}$.

Question

I already solved the question myself, but there is something bottering me. In the exersice is told "Solve, by easy estimations". I couldnt find a boundary for $n$ or something. I started with $a^n = 8 + b^n$ and then follows $a \leq \sqrt[n]{8} + b$, so $a-b \leq \sqrt[n]{8}$. Knowing that $a$ and $b$ in $\mathbb{Z}$ we find that for $n > 3 \implies a-b \leq 1$. But now $a$ and $b$ can be negative aswell. So it isn't a boundary yet. Hopefully help comes soon, otherwise i cant sleep..