If you set $m=n$ you have infinitely many solutions $x=y$.

If $m=n=2$, the solutions are $x=y,x= -y-1$.

If $m=n=3$ we have the factorization $(-x + y) \cdot (x^2 + x*y + y^2 + x + y)$. Maybe for deep reasons, the quadratic factor doesn't have solutions over the integers, but we believe it has infinitely many solutions over $\mathbb{Z}[i]$.

For $m=n=4$ we have a cubic factor, which is expected to be of genus $1$, and according to sage it is genus zero, which might give integral points.

For $ m=n < 30$, the higher degree factor is genus zero, probably some algebraic geometer will explain it.

EDIT This answer doesn't satisfy OP's assumptions. Currently keeping it because of existence of infinitely many solutions over extensions of the integers and the relation with N-conjecture.

If you set $m=n$ you have infinitely many solutions $x=y$.

If $m=n=2$, the solutions are $x=y,x= -y-1$.

If $m=n=3$ we have the factorization $(-x + y) \cdot (x^2 + x*y + y^2 + x + y)$. Maybe for deep reasons, the quadratic factor doesn't have solutions over the integers, but we believe it has infinitely many solutions over $\mathbb{Z}[i]$.

For $m=n=4$ we have a cubic factor, which is expected to be of genus $1$, and according to sage it is genus zero, which might give integral points.

For $ m=n < 30$, the higher degree factor is genus zero, probably some algebraic geometer will explain it.