Note that .$n^5+n^4+1=(n^3-n+1)(n^2+n+1)$. (Using root of unity will help to see easier).

Observe that \[ n^5 + n^4 +1 = \left(1 + n + n^2\right)\left(1 - n + n^3\right). \]Now, let $d$ be the g.c.d. of the factors above. From here, it is not hard to get $d\in\{1,7\}$. Next, since $n^5+n^4+1$ has six positive integer divisors, it is either of form $p^5$ or of form $pq^2$, where $p,q$ are primes. Notice that the former case is impossible as $d\in\{1,7\}$ and $n>1$ clearly. Hence, it is of form $pq^2$ for some primes. If $d=1$, then $n^2<n^2+n+1<(n+1)^2$ yield $n^3-n+1=q^2$, concluding the proof. Hence, assume $d=7$. We then have $n^2+n+1 = 7$ necessarily. From here, $n=2$ is the only possibility, which clearly does not work. Hence $n^3-n+1=q^2$ where $q$ is a prime.