Let $P(m,n)$ denote the assertion, note that $m + f(n) \mid f(m)^2 + mn$, $P(1,1)$ gives size contradiction if $f(1) > 1$ so $f(1)=1$ and $P(1,n) \implies 1 + f(n) \mid 1+ n \implies f(a)=a$ for $a = \text{prime} - 1$, finally note that $P(m,a) \implies m + a \mid f(m)^2 - a^2 \implies m + a \mid f(m)^2 - m^2$, blow $a$ up and you have $f(m)=m$.

$P(a,a)\Longrightarrow a+f(a)\mid f(a)^2-af(a)\Longrightarrow f(a)+a\mid 2af(a)$. $a=1$ gives $f(1)=1$, $P(1,p-1)\Longrightarrow 1+f(p-1)\mid p\Longrightarrow f(p-1)=p-1$ $P(m,p-1)\Longrightarrow m+p-1\mid f(m)^2-(p-1)^2$, but $m+p-1\mid m^2-(p-1)^2\Longrightarrow m+p-1\mid f(m)^2-m^2\Longrightarrow f(m)=m$.

$m+f(n) | f(m)^2-nf(n)$ $P(1,1)$ $1+f(1) |f(1)^2-f(1)$ $1+f(1) | f(1)-1 $ and that implies $f(1)=1$ $P(1,n)$ $1+f(n) |nf(n)-1$ FACT1.$1+f(n) |n+1$ so that $n \geq f(n)$ Let's take any prime $p$ $P(p,p)$ $p+f(p) | f(p)^2-pf(p)$ $\gcd(f(p),p)=1$ $\implies$ $f(p)=p$ $\gcd(f(p),p)=p$ $\implies$ $f(p)=p$ by FACT1. $P(p,m)$ $p+f(m)|p^2-mf(m)$ $p+f(m) |p^2+pm$ $p+f(m) | p(m-f(m))$ $p+f(m) | f(m)(m-f(m)$ if we take p sufficiently large we get contradicition so that $m=f(m)$ is indeed and unique solution.

I just realized that this is in $\mathbb{N}$ but a cool problem though , i will not post the correct solution $\textbf{Answer:}$ $f(n)=n \forall n \in \mathbb{N}$. Clearly this works $\textbf{Solution:}$ Let $Q(m,n)$-denote the given asseriton. First we claim that $f(0)=0$ Proof: $Q(0.1)\implies f(1) \mid f(0)^2-f(1) \implies f(1) \mid f(0)^2-f(1)+f(1)=f(0)^2 \implies f(1) \mid f(0)^2 \implies f(1)^2 \mid f(0)^4$ $Q(0,1) \implies 1+f(0) \mid f(1)^2 \mid f(0)^4 \implies 1+f(0) \mid f(0)^4 \implies 1+f(0) \mid f(0)^4-f(0)^3(1+f(0))=f(0)^4-f(0)^3-f(0)^4=-f(0)^3 \implies 1+f(0) \mid -f(0)^3 \implies 1+f(0) \mid -f(0)^3+f(0)^2(1+f(0))=-f(0)^3+f(0)^2+f(0)^3=f(0)^2 \implies 1+f(0) \mid f(0)^2 \implies 1+f(0)\mid f(0)^2-f(0)(1+f(0))=f(0)^2-f(0)-f(0)^2=-f(0) \implies 1+f(0) \mid -f(0) \implies 1+f(0) \mid -f(0)+1(1+f(0))=-f(0)+1+f(0)=1 \implies 1+f(0) \mid 1 \implies 1+f(0)=1 \implies f(0)=0$ Let $p$-be a prime number. $Q(p,0) \implies p \mid f(p)^2 \implies f(p)=p$ $Q(p,n) \implies p+f(n) \mid p^2-nf(n)$ $Q(n,p) \implies n+p \mid f(n)^2-p^2=(f(n)-p)(f(n)+p) \mid p^2-nf(n) \implies n+p \mid p^2-nf(n) \implies n+p \mid p^2-nf(n)-(n+p)(n-p)=p^2-nf(n)+n^2-p^2=n^2-nf(n) \implies n+p \mid n^2-nf(n)$ Now if we pick $p$ large we get that the $RHS$ is zero so $n^2-nf(n)=0 \implies f(n)=n \forall n \in \mathbb{N}$. As desired $\blacksquare$

$ Q(n,p)$ $ \implies n+p \mid f(n)^2-p^2$ $=(f(n)-p)(f(n)+p) \mid p^2-nf(n)$ $ \implies n+p \mid p^2-nf(n)$ $ \implies n+p \mid p^2-nf(n)-(n+p)(n-p)$ $=p^2-nf(n)+n^2-p^2=n^2-nf(n)$ $\implies n+p \mid n^2-nf(n) $