We have that \[ \min \{f(bc),f(ca),f(ab)\} \le f(ab+bc+ca) \le \max \{f(bc),f(ca),f(ab)\} \]for all $a,b,c\in \mathbb{Z}^{+}$. By putting $b=c=1$, we obtain \[ \min \{f(a),f(1)\} \le f(2a+1) \le \max \{f(a),f(1)\}. \]Furthermore, if we assume $f(a)\neq f(1)$ for some $a\in \mathbb{Z}^{+}$, then we have the strict inequality \[ \min \{f(a),f(1)\} < f(2a+1) < \max \{f(a),f(1)\}. \]By considering iterations, this tells us that $\{f(2^na+2^n-1)\}_{n\ge 0}$ sequence is either strictly increasing and bounded above by the constant $f(1)$, or strictly decreasing and bounded below by the constant $f(1)$. However, no such sequence exists in $\mathbb{Z}^{+}$. Therefore, we should have $f(a)=f(1)$ for all $a\in \mathbb{Z}^{+}$.

We claim all such functions are constant, which obviously works. Let $2<x$ be an integer. Let $f(1)=a$ and $f(2^kx-1)=a_k$. Notice $$2f(2^kx-1)+(2^kx-1)f(1)=(2^kx+1)f(2^{k+1}x-1)\iff a_{k+1}=\frac{2a_k+(2^kx-1)a}{2^kx+1}$$Notice if any $a_k$ is equal to $a$ then they all are so assume none of them are. For $a_{k+1}$ to be an integer we must have $2^kx+1|2(a_k-a)$. So there must exist an integer $K$ for which for all $K\leq k$ we have $a<a_k$. However then for all $K\leq k$, $a_k$ is a strictly decreasing function, a contradiction. So $a=a_0=f(x-1)$ finishing the problem.