Suppose that $\limsup_{r \rightarrow \infty}\frac{M_f(r)}{r^n}<\infty$.
This implies that there exists a constant $M$ such that for all $r$ sufficiently large, $|f(re^{i\theta})| \leq M r^n$ for every $\theta$.
Write $f(z)= \sum_{k=0}^{\infty}a_kz^k$. Fix $r$ large. By Taylor's theorem, $a_k=\frac{1}{2\pi i}\int_{\{|z|=r\}} \frac{f(z)}{z^{k+1}}dz$ and thus $|a_k| \leq \frac{2\pi r Mr^n}{2\pi r^{k+1}} = Mr^{n-k}.$ This holds for all $r$ sufficiently large, so if $k>n$, we find $a_k=0$, by letting $r \rightarrow \infty$. Thus, $f$ is a polynomial of degree at most $n$. Since $f$ is not identically zero, then $f$ has a finite number of zeros.