Croatian Mathematical Society Competitions

Note that $n^5 + n^4 + 1 = (n^3 - n + 1)(n^2 + n + 1)$, and that
$$\begin{align*} d &= \gcd(n^3 - n + 1, n^2 + n + 1) \\ &= \gcd(n^2 + n + 1, -n^2 - 2n + 1) \\ &= \gcd(n^2 + n + 1, n - 2) \\ &= \gcd(n - 2, 7), \end{align*}$$
hence we have two cases:

1) $d = 7$

This implies $7 \mid m$ and $7 \nmid n$, from which we get $7 \mid m - 7n$ and $7 \nmid m - 4n$.
Therefore, there is no solution in this case.

2) $d = 1$

This implies that $n^3 - n + 1$ and $n^2 + n + 1$ are both squares of integers. That is true only for $n = 0$ and $n = -1$, since $n^2 < n^2 + n + 1 < (n + 1)^2$ holds for $n \ge 1$, and $(n+1)^2 < n^2 + n + 1 < n^2$ holds for $n < -1$.
Both $n = 0$ and $n = -1$ yield $m^2 = 1$, i.e. $m = \pm 1$, and among four possibilities only two satisfy the given conditions: $(m, n) = (-1, 0)$ and $(m, n) = (1, 0)$.