Solution:

For an arbitrary positive integer $n$ we choose $m$ such that $f(n) \mid m$ holds. Then it follows that $f(n) \mid m^{2}+f(n)$ and $f(n) \mid m f(m)+n$. Since $m$ is divisible by $f(n)$, $n$ must also be divisible by $f(n)$. Thus we have $f(n) \mid n$, which means $f(n) \leq n$ for all $n \in \mathbb{Z}^{+}$.

In particular this yields $f(1) \mid 1$, hence $f(1)=1$.

We further assume that there exists an $m \in \mathbb{Z}^{+}$ with $f(m)<m$. With $n=1$ we obtain $m^{2}+1 \mid m f(m)+1$, but on the other hand we then have $0<m f(m)+1<m^{2}+1$, so that $m^{2}+1$ cannot divide $m f(m)+1$ - contradiction! Hence $f(m) \geq m$ holds, and combined with the above it follows that $f(n)=n$ for all $n \in \mathbb{Z}^{+}$.

Substituting confirms via $n^{2}+n=n^{2}+f(n) \mid n f(n)+n=n^{2}+n$ that this function is indeed - therefore - the only solution.