The $x_i$ are either all equal to $1 + \frac{2m}{n} = \frac{(m+1)^2}{m^2+1}$ or exactly one is equal to $m+1$ and the other are all equal to $1 + \frac{1}{m}$. The verification offers no difficulty and is hence omitted.

Leaving aside the trivial case where the $x_i$ are all equal, consider a solution $x_1, x_2, \dots, x_n$ whose entries are not all equal. Let $s = x_1^2 + x_2^2 + \dots + x_n^2$ and notice that each $x_i$ is a root of the quadratic polynomial $2mX^2 - sX + s$. Since the $x_i$ are not all equal, and each $x_i$ is positive (in fact, at least $1$), the roots $u$ and $v$ of this quadratic polynomial are distinct positive real numbers satisfying $2muv = s$.
Let $k$ be the number of indices $i$ such that $x_i = u$, so $x_i = v$ for the remaining $n-k$ indices. We may and will assume that $k \ge \frac{n}{2}$; and since the $x_i$ are not all equal, $k \le n-1 = m^2$.
Write $2muv = s = ku^2 + (n-k)v^2 \ge 2uv\sqrt{k(n-k)}$, so $k(n-k) \le m^2$, and recall that $k \ge \frac{n}{2}$, to infer that $k \ge \frac{1}{2}(n + \sqrt{n^2 - 4m^2}) = m^2$. Further, the condition $k \le m^2$ forces $k = m^2$ which in turn forces $v = mu$, by the preceding.
Finally, since the $x_i$ add up to $n + 2m = (m+1)^2$, it follows that $u = 1 + \frac{1}{m}$ and $v = m + 1$, and the solution has the form stated in the first paragraph.