Suppose that the image of integer numbers under the polynomial $P(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0$ contains an infinite geometric progression with common ratio $a \in \mathbb{Z} - \{0\}$. For each $b \in \mathbb{Z}$ we have
$$P(ax + b) = a_n a^n x^n + (n a_n a^{n-1} b + a_{n-1} a^{n-1}) x^{n-1} + \dots$$
$$a^n P(x) = a_n a^n x^n + a_{n-1} a^n x^{n-1} + \dots + a_0 a^n.$$
We can find some $b_1, b_2 \in \mathbb{Z}$ and $N \in \mathbb{N}$ such that for every $x \ge N$,
$$P(ax + b_2) < a^n P(x) < P(ax + b_1),$$
and for every $x \le -N$,
$$P(ax + b_2) < a^n P(x) < P(ax + b_1), \ \text{or } P(ax + b_1) < a^n P(x) < P(ax + b_2).$$
For each $P(x)$ in the geometric progression, $aP(x), a^2P(x), \dots, a^nP(x), \dots$ are all in $P(\mathbb{Z})$, hence there exists some $y \in \mathbb{Z}$ such that $a^n P(x) = P(y)$. If $|x|$ is sufficiently large (there are infinitely many values of $P(x)$ in the geometric progression), using the above inequalities, $ax + b_1 < y < ax + b_2$. Therefore, $y - ax$ is a constant value between $b_1$ and $b_2$. Hence there exists some constant number $c$ such that the equation $a^n P(x) = P(ax + c)$ has infinitely many solutions and consequently $P(ax + c)$ and $a^n P(x)$ are two equal polynomials.
Now, our goal is to find all polynomials $P(x) \in \mathbb{Z}[x]$ satisfying the equation $a^n P(x) = P(ax + c)$. Let $Q(x) = ax + c$. If $\alpha$ is a root of $P(x)$, setting $x = \alpha$ in the equation implies that $P(Q(\alpha)) = 0$ and hence $Q(\alpha), Q^2(\alpha), \dots$ are all roots of $P$. Since $P(x)$ has a finite number of roots, there are some natural numbers $m_1 > m_2$ such that $Q^{m_1}(\alpha) = Q^{m_2}(\alpha)$. This implies $Q^{m_1-m_2}(\alpha) = \alpha$, since $Q$ is injective. Note that if $\beta = \frac{c}{1-a}$, then $Q(\beta) = \beta$ and hence $Q^{m_1-m_2}(\beta) = \beta$. On the other hand, $Q^{m_1-m_2}(x) - x$ is a linear polynomial. Therefore, it has at most one root, hence $\alpha = \frac{c}{1-a}$ is the only root of $P(x)$. Consequently, $P(x)$ has the form $r(x - \frac{p}{q})^n = \frac{r}{q^n}(qx - p)^n$, where $p, q$ and $r$ are three integers such that $(p, q) = 1$.
$P(x) \in \mathbb{Z}[x]$, hence
$$\frac{r}{q^n} p^n \in \mathbb{Z} \Rightarrow q^n \mid r p^n \overset{(q,p)=1}{\Rightarrow} q^n \mid r \Rightarrow \exists s \in \mathbb{Z}; r = q^n s.$$
Therefore, we have $P(x) = s(qx - p)^n$, for some $p, q, s \in \mathbb{Z}$ that $(p, q) = 1$.
We claim that each polynomial of this form satisfies the problem's condition. It is enough to show that the polynomial $qx - p$ satisfies the property. This is equivalent to showing the existence of a geometric progression whose elements are all congruent to $-p$ modulo $q$. Obviously, $\{-p(q+1)^m\}_{m\ge 0}$ is an example of such progression and this completes our proof.