AceXKilla wrote:

Since $|f(z)|$ has a pole at infinity then is a polynomial, but I can't conclude that $f$ is rational, how to proceed?

You're confusing many things together. The correct statement is, a holomorphic function with a pole at $\infty$ is a polynomial. Notice that, $|f(z)|$ is a real-valued function and not holomorphic.

Note that, $\lim_{z \to \infty} |f(z)|=\infty$ implies that $f$ has a pole at $\infty.$

Sketch: The point is that any such functions gives rise to a meromorphic map from the Riemann Sphere $\mathbb{C}_{\infty}$ into itself. Using the compactness of $\mathbb{C}_{\infty},$ you should verify that the number of zeros and poles are finite. Divide $f$ by product of all terms like, $(z-z_o)^k$ where $z_0$ is a zero or pole of order $k$ to kill off all the zeros and poles of $f$ and use the compactness and maximal modulus principal to show that the resulting holomorphic function is constant.

More generally, it's possible to show that any meromorphic function on a compact Riemann surface is a rational function.