Can you find them all?

For $\dfrac {m^2-n}{n^2+m}$ to be an integer,

$\begin{aligned} |m^2-n| & \ge n^2 + m & \small \color{#3D99F6} \text{for } m, n > 0 \text{ and } m^2 - n \ge 0 \\ m^2 - n & \ge n^2 + m \\ m^2 - m - n^2 - n & \ge 0 & \small \color{#3D99F6} \text{A quadratic equation of }m \\ \implies m & = \frac {1 \pm \sqrt{1+4n^2+4n}}2 \\ & = \frac {1 \pm (2n+1)}2 \\ \implies m & = n+1 & \small \color{#3D99F6} \text{for } m>0 \end{aligned}$

Putting $m=n+1$ , then we have $\dfrac {m^2-n}{n^2+m} =$ $\dfrac {(n+1)^2-n}{n^2+n+1} =$ $\dfrac {n^2+2n+1-n}{n^2+n+1} =$ $\dfrac {n^2+n+1}{n^2+n+1} = 1$ . Therefore, $(n+1,n)$ is a solution for any $n$ and there are infinitely many .

Given a positive integer $m$ , let $n = m^2$ . Then $\frac{m^2 - n}{n^2 + m} = 0.$ Thus, there are infinitely many pairs $(m,n)$ .

Put $n=1$ in the given expression we will get

$\dfrac{m^2-1}{1+m}=\dfrac{(m+1)(m-1)}{m+1}=m-1$

Here $m$ can be ant positive integer.SO, there are infinitely many solutions