If one of the two is $0$, the other must be a perfect square. Otherwise, the closest square to $m^2$ is $m^2-(2|m|-1)$, so $|n|\geq 2|m|-1$. By symmetry we also have $|n|\geq 2|m|-1\geq 4|n|-3$ which implies $|m|\leq 1$ and thus also $|n|\leq 1$. By inspection the only other possible solutions that we didn't count are $(\pm 1,1)$; the first doesn't work, while the second does. Thus the answer is $(0,k^2),(k^2,0)$ and $(\pm 1,\mp 1)$ where $k$ is an integer.

Or just simple: $m^2+n=a^2$, and suppose $m>=n$ thus $(2m)^2<=(2a)^2<=4m^2+4m<(2m+1)^2$ and the rest is simple.