Just apply Vieta's

Let $f(x)=x^2 + mx + n$ . We have

$\begin{aligned} f(m) &= 2m^2+n &= 0\\ f(n) &= n^2 + mn + n &= 0\\ \Rightarrow n&=-2m^2 \text{ and } n(n+m+1) &= 0\\ \Rightarrow \quad & -2m^2 (-2m^2 + m + 1) &= 0\\ \Rightarrow \quad & 2m^2(m-1)(2m+1) &= 0 \end{aligned}$

Thus, we have 3 solutions for $m$ , which are $m= - \frac {1}{2}, 0, 1$ . From this, we get all solutions are

$(m, n) = (-\dfrac{1}{2}, -\dfrac {1}{2}), (0, 0), (1, -2)$

Thus, there are 3 ordered solutions.