, 2020

Let: $\deg P(x) = m \ge 1$, $\deg Q(x) = n \ge 1$. By taking the degrees of the two members of the given equation we get the equality:
$$\begin{aligned} 3mn = 1 + m + 3n &\Leftrightarrow (m-1)(3n-1) = 2 \\ &\Leftrightarrow m-1=1, 3n-1=2 \quad \text{or } m-1=2, 3n-1=1\; (\text{impossible in } \mathbb{N}^*) \\ &\Leftrightarrow m=2, n=1. \end{aligned}$$
Let $P(x) = ax^2 + bx + c$, $a, b, c \in \mathbb{R}$, $a \neq 0$, $Q(x) = dx + e$, $d, e \in \mathbb{R}$, $d \neq 0$.
Then, from the given relation we have:
$$a((Q(x))^3)^2 + b(Q(x))^3 + c = xP(x)(Q(x))^3, \quad (1)$$
From which we conclude that $Q(x)|_{c=0} \Rightarrow c=0$. Then from (1) we have:
$$\begin{aligned} & a((Q(x))^3)^2 + b(Q(x))^3 - xP(x)(Q(x))^3 = 0 \\ \Leftrightarrow & (Q(x))^3 \left(a(Q(x))^3 + b - xP(x)\right) \\ \Leftrightarrow & \underset{Q(x) \neq 0}{a(Q(x))^3} + b - xP(x) = 0 \\ & xP(x) = a(Q(x))^3 + b \end{aligned} \quad (2)$$
Then from relation (2) we have:
$$\begin{aligned} & x(ax^2 + bx) = a(dx + e)^3 + b \\ \Leftrightarrow & ax^3 + bx^2 = ad^3 x^3 + 3ad^2 e x^2 + 3ad e^2 x + a e^3 + b \\ \Leftrightarrow & a = ad^3,\; b = 3ad^2 e,\; 3ad e^2 = 0,\; a e^3 + b = 0 \\ & \underset{a,d \neq 0}{\Leftrightarrow} d = 1,\; e = 0,\; b = 0,\; a \in \mathbb{R}. \end{aligned}$$
Hence: $P(x) = a x^2$, $Q(x) = x$, $a \in \mathbb{R}$ and it is easy to verify that they satisfy our problem.