Let $f$ be an entire function and suppose that $\infty$ is a pole of order $n$, so that $g(z)=f(\frac{1}{z})$ has a pole of order $n$ at $0$. Then, in a punctured neighborhood of $0$, $g$ has a Laurent series expansion
$$g(z)=\sum_{k=-n}^\infty a_n z^n.$$
Furthermore
$$\lim_{z\to\infty}g(z)=\lim_{z\to\infty} f(\frac{1}{z})=\lim_{z\to 0} f(z)$$
so
$$g(z)=\sum_{k=-n}^0 a_n z^n.$$
Therefore, in a punctured neighborhood of $\infty$, given by $|z|>R$ for some $R>0$ , $f(z)=g(\frac{1}{z})=\sum_{k=0}^n a_{-n} z^{n}$. Thus, $f$ agrees with a polynomial in a big enough set, and by the identity theorem $f$ is a polynomial. The next result follows:
Lemma. If $f$ is an entire function with $\infty$ as a pole, then $f$ is a nonconstant polynomial.
The reciprocal is true: Every nonconstant polynomial has $\infty$ as a pole.
Proposition. Let $f,g$ be entire functions such that $f\circ g$ is a nonconstant polynomial. Then both $f$ and $g$ are polynomials.
Proof. If $f\circ g$ is a nonconstant polynomial then $f\circ g(z)\to \infty$ when $z\to \infty$. Let's suppose that $g$ is not a polynomial. Then $\infty$ is a removable singularity or an essential singularity of $g$. In the former case $g$ is entire in the extended plane, and by Liouville theorem, $g$ is constant, thus $f\circ g$ is also constant, a contradiction. So $\infty$ is an essential singularity of $g$. In this case, by Casorati-Weierstrass theorem, there exists a sequence $s_n\to\infty$ such that $g(s_n)\to a$ where $a$ is any complex number. Then $f\circ g(s_n)\to f(a)\neq \infty$, a contradiction. So, $g$ must be a nonconstant polynomial. Similarly, if $f$ is not a polynomial, and $\infty$ is an essential singularity of $f$ (If it's removable we can argue as above to show $f\circ g$ is constant) then by Casorati-Weierstrass theorem there is some sequence $s_n\to\infty$ such that $f(s_n)\to a'$. Also, by the Fundamental Theorem of Algebra, the equations $g(z)=s_n$ always have solutions, so we get some choice of sequence $r_n\to\infty$ such that $g(r_n)=s_n$, so that $f\circ g(r_n)\to a'$, another contradiction. So $f$ is a polynomial.
Note: The nonconstant condition cannot be removed: You can take any entire function as $g$ and any constant polynomial as $f$ and $f\circ g$ will be a constant polynomial.
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