It is known that if $n$ divides $a - b$, then $a^n - b^n$ is divisible by $n^2$. Now, choose $z \equiv x \pmod{n}$ and $y \equiv t \pmod{n}$. Then
$$x^n + y^n - (z^n + t^n) = (x^n - z^n) + (y^n - t^n) \equiv 0 \pmod{n^2}$$
Thus, for $z, t$ belonging to $\{0, 1, \dots, n-1\}$, we must calculate maximal number of different values of $z^n + t^n \pmod{n^2}$. Then, there are $x, y$ such that $x^n + y^n \equiv z^n + t^n \pmod{n^2}$. These numbers are at most $\frac{n(n+1)}{2}$. Therefore, there are at least $n^2 - \frac{n(n+1)}{2} = \frac{n(n-1)}{2}$ different values for $m$.