It is easy to see that $f^{n+1}(m)\equiv 1\pmod {f^{n}(m)}$. Then notice that $1-1+1=1,$ so by induction, all $k>n+1$ also satisfy $f^{k}(m)\equiv 1\pmod {f^{n}(m)}$. From here, it is easy to see the claim is true.

This is a very old problem that has been used many times by varying the degree of $f$, nevertheless we prove a more general result . Claim. Given a polynomial $f$ with integer coefficients such that $f(0)=f(1)=1$ , then $\{f(n), f(f(n)), \dots\}$ are all pairwise relatively prime for any $n\in\mathbb{Z}$. Proof. It's easy to see $f^{i}(0)=f^{j}(1)=1$, for any $i, j$. Now take any two $p, q$ with $p<q$, then consider $f^{q-p}(x)$, then clearly it can be written in the form $$f^{q-p}(x)=xg(x)+1$$for some $g\in \mathbb{Z[X]}$. In particular $$f^{q-p}(f^{p}(n))=f^{q}(n)=f^{p}(n)g(f^{p}(n))+1$$that is $(f^{p}(n), f^{q}(n))=1$ for any $p\neq q$, as desired.