Clearly, polynomial $P(x)$ has degree $4n$ and has no real root. Thus, any factor of $P(x)$ has even degree. Suppose that $P(x)$ can be expressed as the product of $n + 1$ polynomials with degree greater than 0, so
$$P(x) = P_1(x) \cdot P_2(x) \cdots P_{n+1}(x)$$
then $P_i(x)$ with $i = 1, 2, \dots, n$ has even degree. Since the sum of degrees of $P_i(x)$ is $4n$ then there are at least two polynomials, let say $P_1(x), P_2(x)$ of degree 2.
Since $P(x)$ has the leading coefficient 1, suppose that $P_1(x), P_2(x)$ have leading coefficients 1, which are
$$P_1(x) = x^2 + a x + b, \quad P_2(x) = x^2 + c x + d \text{ where } a, b, c, d \in \mathbb{Z}.$$
Since $P_1(x), P_2(x)$ have no real roots, we have $P_1(x) > 0, P_2(x) > 0$ for all integers $x$.
We have
$$13 = P(1) = P_1(1) \cdot P_2(1) \cdots P_{n+1}(1)$$
and
$$13 = P(6) = P_1(6) \cdot P_2(6) \cdots P_{n+1}(6).$$
From this, at least one of two numbers $P_1(1)$ and $P_2(1)$ equal 1. Without loss of generality, we assume that $P_1(1) = 1$. It follows that $a = -b$, so $P_1(6) = 36 - 5b$. We observe that $36 - 5b > 0$ and cannot be equal to 13 so $36 - 5b = 1$, thus $b = 7, a = -7$. But in this case the polynomial $P_1(x) = x^2 - 7x + 7$ has real root, which is a contradiction.
Therefore, the above assumption is wrong so $P(x)$ cannot be expressed as the product of $(n + 1)$ non-constant polynomials. $\square$