I will prove this for all real polynomials.
Lemma: If $P \in \mathbb{R}[x]$ satisfies $P(a) = b$ for infinitely many pairs of integers $(a, b)$, then $P \in \mathbb{Q}[x]$.
Proof: Enumerate these integer pairs as $(a_1, b_1), (a_2, b_2), \dots$. Suppose $\deg P = n$. Then, consider $Q \in \mathbb{Q}[x]$
with $\deg Q \leq n$ satisfying $Q(a_i) = b_i$ for
$i = 1, 2, \dots, n+1$. Note that this exists by Lagrange Interpolation.
Then $R(x) := P(x) - Q(x) = 0$ at $x = a_1, a_2, \dots, a_{n+1}$, but
$\deg R \leq n$. Hence $R(x) = 0$, so $P(x) = Q(x)$.
Thus, $P \in \mathbb{Q}[x]$. $_\blacksquare$
Now we can apply this lemma to the question to reduce this to the case where $P \in \mathbb{Q}[x]$. This is equivalent to finding all polynomials $P \in \mathbb{Z}[x]$ such that there exist infinitely pairs of integers $(x, y)$ such that $P(x) = 2^y z$ for some fixed constant $z$.
If we let $a$ denote the leading coefficient, $n$ denote the degree of $P$, and let the sequence of satisfying $(x, y)$ be $(x_1, 1), (x_2, 2), \dots$, then observe that $P(x_k) = 2^{k}z \implies x_k = \sqrt[n]{\frac{2^{k}z}{a}}(1+o(1))$.
This implies for large enough $k$, $x_{k+1} - x_k$ is strictly increasing. However, we also have $x_{k+1} - x_k \mid P(x_{k+1}) - P(x_k)$, so $x_{k+1} - x_k$ is eventually of the form $2^m c$, for some $c \mid z$. Since this difference is strictly increasing and $z$ is a constant value, there will exist some $K$ such that $x_{k+1} - x_k \geq 2^{k-K}$ for large enough $k$. Hence, $\frac{x_k}{2^k} \geq C > 0$ for all positive integers $k$ for some real $C$.
However, if $n > 1$, then $x_k = \sqrt[n]{\frac{2^{k}z}{a}}(1+o(1)) \implies \frac{x_k}{2^k} \to 0$ as $k \to \infty$, so there exists large enough $k'$ such that $\frac{x_{k'}}{2^{k'}} < C$, contradicting the above.
Therefore, $n = 1$, contradicting the hypothesis so we're done.
For linear polynomials, if $a = \frac{p}{q}$ for integers $p, q$ with $q > 0$ and $\gcd(p, q) = 1$, $P$ is of the form $a(x+b)$ where $p \in \{\pm 1, \pm 2\}, q \in \mathbb{N}$ and $b \in \mathbb{Z}$ such that $q \mid b$.