Can $7^n+1$ be a perfect cube when $n>1$? $7^n+1$ (when $n>1$) can in principle be a perfect cube, since it sometimes is divisible by $8$, and sometimes leaves $8$ as a remainder when dividied by $9$. But when I tried to find perfect cubes that is form of $7^n+1$, I did not suceed.
Can $7^n+1$ be a perfect cube?

This solution is based on a solution to a similar problem about $2^n + 1$ found here.

No.
Suppose that $7^n + 1 = m^3$ for some positive integers $n$ and $m$. Rearranging, we get $$7^n = m^3 - 1 = (m-1)(m^2 + m + 1).$$
Because $7$ is prime, we know that both $m-1$ and $m^2 + m + 1$ must both be powers of $7$ (including $7^0 = 1$).

*

*If $m-1 = 7^0 = 1$, then $m = 2$ and hence $n = 1$.

*If $m-1 \neq 7^0$, then $7 \mid (m-1)$. This implies $m \equiv 1 \pmod 7$. However, we would get $$m^2 + m + 1 \equiv 1^2 + 1 + 1 \equiv 3 \pmod 7.$$ This contradicts the fact that $m^2 + m + 1$ must be a power of $7$.

So, excluding $n = 1$, there is no other positive integers $n$ such that $7^n + 1$ is a perfect cube.
