To add on to what Makuasi stated an entire function has a pole of order $n$ at $\infty$ iff the function is a polynomial of order $n$. So if $f$ does not have a pole at $\infty$ $f$ is either bounded or has an essential singularity at $\infty$. $f$ cannot be bounded by Liouville's Thm. So $f$ must have an essential singularity at $\infty$. By Casorati–Weierstrass theorem there is a sequence, $(z_{n})$, such that $|z_{n}| \rightarrow \infty$, and $f(z_{n})\rightarrow 0$
I'd like to attempt an answer. I will have no problem with down-votes or with deleting it, since I am not 100% confident in it.
Let $A$ be the annulus $\frac{2}{3} \le |z| \le \frac{3}{2}$. For a sequence $a_n \in \mathbb{C}$, define $f_{a_n}(z) = f(a_n z)$ and $B_{a_n} = \{a_n z : z \in A \}$.
$B_{a_n}$ is an annulus with inner radius $|a_n|\frac{2}{3}$ and outer radius $|a_n|\frac{3}{2}$.
Let $a_n = n$, n positive integer. Let $f_{n_k}$ be the subsequence of $f_n$ guaranteed by the statement. Then either $f_{n_k} \to \infty$ uniformly on $A$ or $f_{n_k} \to g$ uniformly on $A$ for an analytic function $g$.
In the second case, $g$ is analytic so $g(A)$ is bounded. Using uniform convergence, you can see that There is an $M$ such that $f_{n_k}(A) < M$ for all $k$ sufficiently large. This means $f(z) < M$ for all $z \in B_{n_k}$, with $k$ sufficiently large. Mimicking the proof of Liouville's theorem that uses the Cauchy Integral Formula, this shows that $f(z)$ is constant.
In the first case, for any $M$ you can achieve $|f_{n_k}(z)| > M$ for $k$ sufficiently large. If in addition, for $k$ sufficiently large, $B_{n_k} \cap B_{n_{k+1}}$ is non-emtpy, then this shows $f(z) \to \infty$ as $z \to \infty$, and therefore $f$ must be a polynomial.
If on the other hand there are infinitely many $k$ with $B_{n_k} \cap B_{n_{k+1}}$ empty, then we need to look at what happens in these "gaps". Note that the outer radius of $B_{n_k}$ is in this case strictly less than the inner radius of $B_{n_{k+1}}$. This "gap" is an open annulus, but $f$ is analytic on the closure. Suppose such a gap does not contain a zero of $f$. Then the Minimum Modulus Theorem says that the modulus of $f$ in the gap is at least as great as the minimum modulus on the boundary, and so in the gap $f|(z)| > M$.
Therefore if only finitely many gaps contain zeros, we again see that $f(z) \to \infty$ as $z \to \infty$.
Finally, suppose there are infinitely $n_k$ with $B_{n_k} \cap B_{n_{k+1}}$ empty, and infinitely many of these "gaps" contain a zero of $f$. For the $m$-th gap pick a zero $b_m$, with $|b_m| < |b_{m+1}|$ and $|b_m| \to \infty$. These $b_m$ define a new sequence $f_{b_m}$ of functions on $A$, a subsequence of which converges uniformly. But in this case it cannot converge uniformly to $\infty$, because each function in the sequence has a zero at $z = 1$. Then as in the proof of the second case, $f$ must be constant.
Best Answer
Consider $\infty$ as a singularity point. It's known that it's a pole of an entire function $f$ iff $f$ is a polynomial. Let's show it is a pole. Obviously it's not a removable singularity since then $f$ is bounded. If it's an essential singularity, by the Casserti Weirstrass thm we can find an unbounded sequence $a_n\to\infty$ with $f(a_n)\to 0$, contradiction. Therefore it is a pole and $f$ is a polynomial.
This is false - consider $f(z)=\frac{z^2+1}{z}$. If $a_n$ is unbounded, take $a_{n_k}\to\infty$. Since $\infty$ is a pole of $f$, $f(a_{n_k})\to\infty$ and $f(a_n)$ is unbounded. So $f$ maps unbounded sequences to unbounded sequences, but is not a polynomial.
In fact, one can show that the only functions that are holomorphic in the punctured plane and satisfy this property must be rational with a pole at infinity (this is if and only if).