The solutions are $\left(k, k^{2}\right)$ and $\left(k^{2}, k\right)$, with $k \geq 2$.
We have $m n - 1 \mid \left(n^{3} - 1\right) m - n^{2}(m n - 1) = n^{2} - m$, hence
$$m n - 1 \mid m\left(n^{2} - m\right) - (m n - 1) n = n - m^{2}$$
If $n > m^{2}$, then $m n - 1 \leq n - m^{2} \leq n - 1$, so $m n \leq n$, not possible.
If $n = m^{2}$, then obviously $m^{3} - 1 \mid m^{6} - 1$, so all pairs ($m, m^{2}$), $m \geq 2$ are solutions.
If $n < m^{2}$, from $m n - 1 \leq n^{3} - 1$ we obtain that $\sqrt{n} < m \leq n^{2}$. Then $m n - 1 \leq m^{2} - n < m^{2} - 1$, hence $n < m$. If $n^{2} - m > 0$, we get $m n - 1 \leq n^{2} - m < n^{2} - 1$, so $m < n$, a contradiction. It follows $n = m^{2}$, satisfying the condition in the problem since $m^{3} - 1 \mid m^{3} - 1$, so all pairs $\left(n^{2}, n\right), n \geq 2$, are also solutions.