For $m=0, x_1 = 1+2m = 1$ is the only solution. If $m > 0,$ let $c = \frac{2m}{x_1^2 + \dots + x_n^2} > 0, f(x) = \frac{x-1}{x^2}.$ Then $f(x_i)=c$ for all $x_i$ ($x_i = 0$ is impossible because plugging it in gives $0 = 1$). But $f$ is strictly decreasing with asymptotes $x = \pm 1, y=0,$ so $f(x) = c$ has at most $2$ solutions. Thus, there are at most $2$ distinct $x_i.$ Let $x_1 = \dots = x_k = r, x_{k+1} = \dots = x_n = s.$ Then $f(r) = f(s) = \frac{2m}{kr^2+(n-k)s^2},$ which we can solve for $r,s$ given any value of $k, 0 \le k \le n.$ This gives $\left\lfloor \frac{n+1}{2}\right \rfloor$ solutions since $k, n-k$ are the same upon swapping $r, s.$

As in the above, let $c = \frac{2m}{x_1^2 + \dots + x_n^2}$. Then, $c > 0$, and thus $x_i \geq 1$ for all $1 \leq i \leq n$. Note that each $x_i$ is a root of the equation $cx^2 - x + 1 = 0$ in $x$, so there are only at most 2 possible values for the $x_i$s, say $a$ and $b$, with $a \neq b$ and $a, b > 1$. If all $x_i$s are equal, direct computation gives us $\boxed{x_i = \frac{(m+1)^2}{m^2 + 1}}$ for all $1 \leq i \leq m^2 + 1 = n$. Otherwise, let $k_1$ be the number of indices $i$ with $x_i = a$, and $k_2 = n - k_1$ be the number of indices $i$ with $x_i = b$. Without loss of generality, suppose $k_1 \leq k_2$. Now, summing up all the equations for $i = 1, 2, \ldots, n$ gives us \begin{align*} \sum_{i = 1}^n x_i &= n + 2m \\ k_1 a + k_2 b &= (m + 1)^2 \\ k_1 (a - 1) + k_2 (b - 1) &= (m + 1)^2 - n \\ &= 2m \end{align*}Meanwhile, since $a$ and $b$ are roots of $cx^2 - x + 1 = 0$, we have $ab = a + b = \frac{1}{c}$, so $(a - 1)(b - 1) = 1$. Hence, AM-GM inequality yields us $$ k_1 (a - 1) + k_2 (b - 1) \geq 2 \sqrt{k_1 k_2} \geq 2m $$where equality holds if and only if $k_1 (a - 1) = k_2 (b - 1)$ for the left inequality, and $k_1 = 1, k_2 = m^2$ for the right inequality. So, we obtained $a - 1 = m^2 (b - 1)$. Since $(a - 1)(b - 1) = 1$, then $a = m + 1$ and $b = \frac{1}{m} + 1$. Thus, we obtain $\boxed{\left(m + 1, \frac{1}{m} + 1, \frac{1}{m} + 1, \ldots, \frac{1}{m} + 1\right)}$ and its permutations. Check that $x_1^2 + x_2^2 + \ldots + x_n^2 = 2(m + 1)^2$ and the rest is a matter of plugin.

That solution is surprising. I mistakenly thought I had gotten through the hard part of the problem, so I left the rest and moved onto a different problem.

2016 Canada P2