Suppose that there exists such polynomial $P(x)$. Let $Q(x) = P(1+x) - 1$ then $Q(x) \in \mathbb{Z}[x]$. We have $Q(\sqrt[3]{2}) = \sqrt[3]{2}$ and $Q(\sqrt{5}) = 1 + 3\sqrt{5}$.

Therefore, $Q(x) - x$ has an irrational root $\sqrt[3]{2}$. Since $x^3 - 2$ is irreducible over $\mathbb{Z}[x]$ and has the same root $\sqrt[3]{2}$, we get $x^3 - 2 \mid Q(x) - x$. Then there exists $R(x)$ with integer coefficients such that
$$Q(x) - x = (x^3 - 2)R(x).$$
Because $R(x) \in \mathbb{Z}[x]$ so $R(\sqrt{5}) = a + b\sqrt{5}$ for some $a, b \in \mathbb{Z}$. Let $x = \sqrt{5}$ then:
$$1 + 3\sqrt{5} = (5\sqrt{5} - 2)(a + b\sqrt{5}) = 25b - 2a + (5a - 2b)\sqrt{5},$$
This implies $5a - 2b = 3$ and $2a = 25b - 1$, which has not a pair integer $(a,b)$ satisfying, a contradiction. So the answer is no.