SAMC

The solutions are $\left(k,k^2\right)$ and $\left(k^2,k\right)$, with $k\ge 2$. We have $mn-1\mid \left(n^3-1\right)m-n^2(mn-1)=n^2-m$, hence $$mn-1\mid m\left(n^2-m\right)-(mn-1)n=n-m^2$$ If $n>m^2$, then $mn-1\le n-m^2\le n-1$, so $mn\le n$, not possible. If $n=m^2$, then obviously $m^3-1\mid m^6-1$, so all pairs ($m,m^2$), $m\ge 2$ are solutions. If $n<m^2$, from $mn-1\le n^3-1$ we obtain that $\sqrt{n}<m\le n^2$. Then $mn-1\le m^2-n<m^2-1$, hence $n<m$. If $n^2-m>0$, we get $mn-1\le n^2-m<n^2-1$, 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 $\left(n^2,n\right),n\ge 2$, are also solutions.