Find all functions over $\mathbb{Z}^+$ satisfying this divisibility criterion

Question

The problem is to find all functions $f: \mathbb{Z}^+\rightarrow \mathbb{Z}^+$ such that $[f(m)]^2 + f(n)$ divides $(m^2+n)^2$ for all $m$,$n\in\mathbb{Z}^+$.

I already have my claim and attempted proof, I just want to know whether it's good enough:

First, let $g = g(m,n) = [f(m)]^2 + f(n)$. Considering the case $m = 1 = n$, then $g | (1^2 + 1)^2 = 4$. If $f(1) = 2$, then $g = 5$, a contradiction. Clearly, $f(1)$ cannot be greater than 2, so we must have that $$f(1) = 1$$ Next, let $m = 1$ and $n= p - 1$ with $p$ a prime number. Then $g = f(p-1) + 1$ divides $p^2$. Subsequently, $g$ must divide $p$ itself. For $p = 2$, $g|2$ if and only if $f(1) = 1$, which is consistent with the earlier result. For $p = 3$, $g|3$ if and only if $f(2) = 2$. In general, $$f(p-1) + 1 | p$$ for all $p$, provided $f(p-1) = p-1$. In other words, the solution is: $$f(x) = x$$

Answer 1

Here are a few points of improvement for your attempted proof:

1. The notation $g=g(m,n)$ is rather confusing because $g$ depends on $m$ and $n$. You write $$g\mid(1^2+1)^2=4\qquad\text{ and later }\qquad g=f(p-1)+1,$$ but these two instances of $g$ do not refer to the same thing. Write $g(1,1)$ and $g(1,p-1)$ instead.
2. You have shown that $f(p-1)+1$ divides $p^2$. Because $f(p-1)>0$, that means either $f(p-1)=p-1$ or $f(p-1)=p^2-1$. It is not clear how you conclude that $f(p-1)=p-1$.
3. From the claim that $f(p-1)=p-1$ for every prime number, you conclude that $f(x)=x$ for every positive integer. It is not clear how this should follow.

Given these two gaps in your attempted proof, I'd say your proof is not good enough.