First, we will show that $Q_1 = P(x^2)$ is irreducible over $\mathbb{Z}$. Let $\deg P = n$. Then $\deg Q_1 = 2n$. To the contrary, assume that there exist $R_1(x), R_2(x) \in \mathbb{Z}[x]$ such that $Q_1(x) = R_1(x)R_2(x)$. It is clear that $Q_1(x) = Q_1(-x) = R_1(-x) \cdot R_2(-x)$. Let $F(x)$ be common factor of $R_1(x)$ and $R_1(-x)$, with maximum degree. Then it is clear that $F(x) = F(-x)$. Thus, $F(x) = G(x^2)$ for some $G(x)$. It is also clear that $\deg G \le n-1$. From $G(x^2) \mid Q(x) = P(x^2)$, we can conclude that $G(x) \mid P(x)$. This means that $\deg G = 0$ and $G(x) \equiv c$. Since $R_1(x)R_2(x) = R_1(-x)R_2(-x)$ we get
$$R_1(-x) \mid R_2(x), \quad R_2(-x) \mid R_1(x).$$
Thus $\deg R_1 = \deg R_2 = n$, $R_2(x) = cR(-x)$ and $Q(x) = cR_1(x)R_1(-x)$. Since $Q(x)$ is a unitary polynomial $c = (-1)^n$. Hence $|P(0)| = |(-1)^n(R_1(0))| = a^2, a \in \mathbb{Z}$. But given is $|P(0)| = 2010$, which leads to contradiction. Thus $P(x^2)$ is irreducible over $\mathbb{Z}$. Now we define a sequence of polynomials by the following recurrent relation:
$$Q_{n+1}(x) = Q_n(x^2), n = 1, 2, \dots$$
It is obvious that $|Q_n(0)| = 2010$. Thus $Q_n(x)$ is irreducible over $\mathbb{Z}$.