That seems like too little information!

A more elementary proof, using lower levels of complex analysis, runs as follows. We can find $R > 0$ such that $|f(z)| \ge 1$ for all $|z| \ge R$ . Thus all zeros of $f$ lie in the compact set $\{z \,:\, |z| \le R\}$ . Since $f$ is not constant, it must have at most a finite number of zeros, and so we can write $f(z) \; = \; p(z)g(z)$ where $p(z)$ is a polynomial and $g(z)$ is an entire function with no zeros. But then $g^{-1}$ is entire without zeros, and we can find $K > 0$ such that $\left|\tfrac{1}{g(z)}\right| \; = \; \left| \tfrac{p(z)}{f(z)}\right| \; \le \; K|z|^{\partial p} \hspace{1cm} |z| \ge R$ where $\partial p$ is the degree of $p(z)$ . Considering Taylor's Theorem for the entire function $g^{-1}$ we know that $\tfrac{1}{g(z)} \; = \; \sum_{m=0}^\infty b_m z^m \;, \hspace{1cm} z \in \mathbb{C}$ where $b_m \; = \; \frac{1}{2\pi i}\int_{|z|=S} \frac{dz}{g(z)z^{m+1}} \hspace{1cm} m \ge 0\,,\, S > 0$ and hence it follows that $\left|b_m\right| \; \le \; KS^{\partial p-m} \hspace{1cm} m \ge 0 \,,\, S \ge R$ and hence that $b_m = 0$ for all $m > \partial p$ . Thus $g^{-1}$ is a polynomial without zeros, and hence is constant. This implies that $f$ is a polynomial . This implies that $f$ is surjective, and that the range of $f$ is cofinite, but does not force $f$ to be either injective or bijective.

Note that hypothesis is just that $f$ can be analytically continued to a meromorphic function $g$ on ${\overline {\mathbb {C} }}$ (the Riemann Sphere) with $\infty$ being the only pole (note that this extension is unique).

This means that $g(\frac{1}{z})$ has a pole at zero. This means that $g$ looks like the following:

$g(z) = P(z) + h(z)$

for some polynomial $P$ and some $h$ holomorphic on ${\overline {\mathbb {C}}}$ such that $g(0) = P(0)$

In particular, $g - P$ is a bounded entire function that vanishes at a point.

By Liouville's theorem, $g = P$ on ${\mathbb {C}}$

Hence, $f = P$ on ${\mathbb {C}}$

P.S.: Note that the converse is trivial. Hence, only polynomials have this property.