Let $P(x)$ and $Q(x)$ be polynomials with real coefficients such that for infinitely many positive integers $n$, $P(1)P(2)\dots P(n) = Q(n!)$.

Let $d$ be the degree of $P(x)$ and $e$ the degree of $Q(x)$.

For large $n$, $P(1)P(2)\dots P(n)$ is a product of $n$ terms, each of degree $d$, so the degree of $P(1)P(2)\dots P(n)$ as a polynomial in $n$ is $d n$ (since each $P(k)$ is a degree $d$ polynomial in $k$, and the product over $k = 1$ to $n$ gives degree $d n$ in $n$).

On the other hand, $Q(n!)$ is a polynomial evaluated at $n!$, so as $n$ grows, $n!$ grows very rapidly. The degree of $Q(n!)$ as a function of $n$ is $e$ times the degree of $n!$ as a function of $n$, but $n!$ is not a polynomial in $n$.

But for the equality to hold for infinitely many $n$, the growth rates must match. Consider the leading behavior:

Let $P(x) = a_d x^d + \dots$ and $Q(x) = b_e x^e + \dots$.

Then for large $n$:
$$P(1)P(2)\dots P(n) \sim a_d^n (1 \cdot 2 \cdot \dots \cdot n)^d = a_d^n (n!)^d$$

So $P(1)P(2)\dots P(n) \sim a_d^n (n!)^d$.

On the other hand,
$$Q(n!) \sim b_e (n!)^e$$

So for the equality to hold for infinitely many $n$, we must have $e = d$ and $a_d^n = b_e$ for all $n$ large enough, which is only possible if $a_d = 1$ and $b_e = 1$ (since $a_d^n$ is exponential in $n$ unless $a_d = 1$ or $a_d = 0$).

But if $a_d = 1$, then $P(x)$ must be $x^d$ (since otherwise the lower degree terms will affect the product for large $n$), and $Q(x) = x^d$.

Let us check this:
If $P(x) = x^d$, then $P(1)P(2)\dots P(n) = (1^d)(2^d)\dots(n^d) = (1 \cdot 2 \cdot \dots \cdot n)^d = (n!)^d$.

If $Q(x) = x^d$, then $Q(n!) = (n!)^d$.

So $P(1)P(2)\dots P(n) = Q(n!)$ for all $n$.

Now, suppose $P(x)$ is not a monomial. Then for large $n$, $P(k) \sim a_d k^d$, so $P(1)P(2)\dots P(n) \sim a_d^n (n!)^d$, but the lower degree terms will affect the product by a factor that is not a polynomial in $n!$, so the equality cannot hold for infinitely many $n$.

Therefore, the only solutions are:
$$P(x) = x^d, \quad Q(x) = x^d, \quad \text{for some integer } d \ge 0.$$

Thus, all polynomials $P(x) = x^d$ and $Q(x) = x^d$ for $d \ge 0$ are solutions.