$ (m,n)=(k,k^2)$ or $ (m,n)=(k^2,k)$.

can anyone prove that other than general solution (m,n)=(k,k^20,there is no other solution?

$ \frac{n^3-1}{mn-1}=1+mn+n^2\frac{n-m^2}{mn-1}=1+mn-n^3+mn^2\frac{n^2-m}{mn-1}$. Because $ (mn-1,n^2)=1=(mn-1,mn^2)$ must be $ a=\frac{n-m^2}{mn-1}\in Z, b=\frac{n^2-m}{mn-1}\in Z$. If $ 1<m<\sqrt n$, then $ 0<a<1$ is not integer. If $ \sqrt n <m\le n>1$ then $ -1<a<0$ is not integer. If $ n<m<n^2$, then $ 0<b<1$ is not integer. If $ n^2<m$, then $ -1<b<0$ is not integer. Therefore $ (m,n)=(k,k^2),(k^2,k)$.

Notice $0 \equiv n^3-1 \equiv n^3-mn (mod mn-1)$, since $gcd(n, mn-1)=1$, then $mn-1 \mid n^2-m$, now it's easy.