Answer: $P(x) = a x^2 - a(b+1)x + a b$, $Q(x) = x^2 - (b+1)x + 2b$, where $a, b \in \mathbb{R}$ and $a \ne 0$.

Denote the degrees of polynomials $P$ and $Q$ by $m$ and $n$, respectively. In the equality
$$P(Q(x)) = P(x)Q(x) - P(x) \quad (1)$$
compare the degrees of both sides:
$$m n = m + n \iff (m-1)(n-1) = 1 \iff m = n = 2.$$
Therefore we can denote $P(x) = \alpha x^2 + \beta x + c$. Hence (1) can be written in the form $\alpha(Q(x))^2 + \beta Q(x) = P(x)Q(x) - (P(x)+c)Q(x)$. Since polynomials $P$ and $Q$ have equal degrees, $P(x)+c = a Q(x)$ for some real number $a \ne 0$. The equality (1) now takes the form
$$a Q(Q(x)) - c = (a Q(x) - c) Q(x) - (a Q(x) - c) = a(Q(x))^2 - (a+c)Q(x) + c.$$
Since $Q$ is nonconstant, it attends infinitely many values, hence $a Q(x) = a x^2 - (a+c)x + 2c$, or equivalently $Q(x) = x^2 - (b+1)x + 2b$ (where $b = \frac{c}{a}$). Then $P(x) = a Q(x) - a b$, i.e. $P(x) = a x^2 - a(b+1)x + a b$. It is easy to verify that such polynomials $P$ and $Q$ satisfy the problem conditions (1). Indeed, since $P$ and $Q$ satisfy $P(x) = a Q(x) - a b$, the equality (1) can be written as $a Q(Q(x)) - a b = (a Q(x) - a b) Q(x) - (a Q(x) - a b)$, or, after simplification, $Q(Q(x)) = (Q(x))^2 - (b+1)Q(x) + 2b$, which is true for $Q(x) = x^2 - (b+1)x + 2b$.