$\textrm{Ans:}$ $f(n)=-n-1$ for all integer $n$. It is clear that it is a solution to the above equation, we now show that it is the only one. First, let $f(n)=-g(n)$, denote the new equation by $Q(m,n)$ \[Q(m,n) : g(m+g(g(n)))=g(g(m+1))+n.\]We have $Q(0,n) \implies g(g(g(n)))=g(g(1))+n$ and we can see that $g$ is injective. Then, $Q(n,0) \implies g(n+1)=n+g(g(0))$ and $g$ is linear so as $f$ is linear, now by plugging $f(n)=an+b$, we see that $\boxed{f(n)=-n-1}$ is the only solution.