First let $m=n,m^2|(m^2+1)\implies m^2|1,m=1$. Then consider $4(mn+1)=m^2+n^2\implies (m-2n)^2-3n^2=1$, a pell equation. Then $4mn+1>(m+n)^2>4mn$, contradiction.

Shu wrote: Find all pairs $(m, n)$ of positive integers for which $4 (mn +1)$ is divisible by $(m + n)^2$. correct solution: Because $4mn\le (m+n)^2\le4mn+4$ to be correct and $mn\ge 4$ we have $(m+n)^2=4mn+4$ or $=4mn$ which only occurs with $m=n.$ Otherwise we have $(m-n)^2=4=2^2.$ So the solutions are $(1,1),(x,x+2),(x+2,x)$.

My idea is same as SCP

$|4mn|\le (m+n)^2\le |4mn+4|$

What about this question? (m+n)^2 is divisible by 4(mn+1).

There are infinite solutions just put n=1 and $m\equiv3 mod4$