Let $f: \mathbb{Z} \rightarrow \mathbb{Z}$ satisfy the given functional equation. Putting $n=0$ in this equation, we get
$$f(2 m+f(m)+f(m) f(0))=m \quad \forall m \in \mathbb{Z} ;$$
therefore, $f$ is surjective, so there exists $u \in \mathbb{Z}$ such that $f(u)=-1$.
With $m=u$, the given equation gives us
$$f(2 u-1-f(n))=-n+u .$$
Now, if $a, b$ are some integers such that $f(a)=f(b)$, then
$$u-a=f(2 u-1-f(a))=f(2 u-1-f(b))=u-b,$$
which implies that $a=b$. Hence, $f$ is also injective.
Next, putting $n=u$, we have
$$f(2 m+f(m)-f(m))=u f(m)+m \Leftrightarrow f(2 m)=u f(m)+m(*)$$
for all $m \in \mathbb{Z}$. In ( $*$ ), letting $m=0$, we see that $f(0)=u f(0)$, so $u=1$ or $f(0)=0$.
If $f(0)=0$, then with $m=u$, (*) would imply $f(2 u)=-u+u=0= f(0)$, i.e. $2 u=0 \Leftrightarrow u=0$ (since $f$ is injective), and thus $f(0)=-1$, a contradiction!
Hence, $u=1 \Leftrightarrow f(1)=-1$ and ( $*$ ) becomes
$$f(2 m)=f(m)+m$$
for all $m \in \mathbb{Z}$. Here, letting $m=1$, we obtain $f(2)=0$.
In the given functional equation, putting $m=n=0$, we have
$$f\left(f(0)+f(0)^2\right)=0 \Leftrightarrow f(0)+f(0)^2=2 \Leftrightarrow f(0)=1 \text { or } f(0)=-2 .$$
If $f(0)=1$, then it would follow from the given equation with $m=0, n=2$ that $f(1)=2$, a contradiction (because $f(1)=-1$ )!
Hence, $f(0)=-2$. In the functional equation, putting $n=0$, we get
$$f(2 m-f(m))=m \quad \forall m \in \mathbb{Z}$$
Using this, we see that if $f(m)=m-2$ then $f(m+2)=m$; but $f(0)= -2, f(1)=-1$, we can easily prove by induction that $f(n)=n-2$ for all $n \geq 0$.
In the given equation, letting $m=1$, we obtain
$$f(2-1-f(n))=-n+1 \Leftrightarrow f(1-f(n))=1-n \quad \forall n \in \mathbb{Z} .$$
Now replacing $n$ by $n+3$ and letting $n \geq-3$, we have
$f(1-f(n+3))=-n-2 \Leftrightarrow f(1-(n+1))=-n-2 \Leftrightarrow f(-n)=-n-2$.
So $f(n)=n-2$ for all $n \in \mathbb{Z}$.
It is easy to check that this function satisfies the given condition.