Expanding both sides, $$m^3+mn+m^2n^2+n^3=m^3-3m^2n+3mn^2-n^3$$ Note that $m^3$ can be canceled and as $n \neq 0$ , $n$ can be factored out.
Writing this as a quadratic equation in $n$ : $$2n^2+(m^2-3m)n+(3m^2+m)=0$$ .
The discriminant $b^2-4ac$ equals $$(m^2-3m)^2-8(3m^2+m)$$ $$=m^4-6m^3-15m^2-8m$$ , which we want to be a perfect square.
Miraculously, this factors as $m(m-8)(m+1)^2$ . This is square iff (if and only if) $m^2-8m$ is square or $m+1=0$ . It can be checked that the only nonzero $m$ that work are $-1, 8, 9$ . Finally, plugging this in and discarding extraneous roots gives all possible ordered pairs $(m, n)$ as $$\{(-1,-1),(8,-10),(9,-6),(9,-21)\}$$ .