Let us rewrite the equation: $n(n^2 - 5n) = m(m^5 - 1)$. If $n \le m$ then $n^2 - 5n \ge m^5 - 1 \ge n^5 - 1$. Contradiction, Thus, $n > m$. Now $n^3 > n^3 - 5n^2 + n > n^3 - 5n^2 + m = m^6$. Therefore, $n^3 > m^6$, equivalently $n > m^2$. Let $n = m^2 + k$ for some $k > 0$. By putting $m^2 = n - k$ to the main equation

we get $5n^2 - 3m^2nk - k^3 - m = 0$. Thus, $n$ divides $k^3 + m$ and $n \le k^3 + m$. Now $5n^2 = k^3 + 3m^2kn + m \ge n(3m^2k + 1)$. Therefore, $5n \ge 3m^2k + 1$ or $5m^2 + 5k \ge 3m^2k + 1$. Since $m^2(3k - 5) \le 5k - 1$ then $m^2 \le \frac{5k-1}{3k-5} = 9 + \frac{44-22k}{3k-5}$. Therefore, if $k > 1$ then $m^2 \le 9$. Thus, $m$ takes $1,2,3$. If $m = 1$ then $n^2(5-n) = 0$ and $n = 5$. If $m = 2$ then $n^2(5-n) = 62$ gives no solution. If $m = 3$ we get $n^3 - 5n^2 - 726 = (n-11)(n^2+6n+66) = 0$ giving the only solution $n = 11$. Now let $k = 1$. Then by putting $n = m^2 + 1$ to the main equation we get $m(3m^3 - 2m + 1) = 4$. Then $m \ne 1$ and for $m > 1$ the left hand side exceeds $4$. Done. Solutions: $(m, n) = (1, 5), (3, 11)$.