All solutions are $$\boxed{\begin{cases}P(x)=ax^2-an^2x,~Q(x)=x+n~~~\text{where }a\in\mathbb R_{\ne0}\text{ and }n\in\mathbb R\\ P(x)=ax^2-an^2x,~Q(x)=-x+n~~~\text{where }a\in\mathbb R_{\ne0}\text{ and }n\in\mathbb R\end{cases}}$$ Proof: Let the degree of $P$ and $Q$ be $p$ and $q$ respectively. We have: $$2pq=p+2q\implies(p-1)(2q-1)=1$$This shows us that $p=2$ and $q=1$. Let $P(x)=ax^2+bx+c$ and $Q(x)=mx+n$. We have: $$a(mx+n)^4+b(mx+n)^2+c=(ax^2+bx+c)(mx+n)^2$$This equation must be true for all real numbers $x$. Substituting $x=-\frac nm$ gives: $$c=0$$Therefore, $P(x)=ax^2+bx$. $$a(mx+n)^4+b(mx+n)^2=(ax^2+bx)(mx+n)^2$$Substituting $x=0$ now gives: $$an^4+bn^2=0$$This shows us that either $n=0$, or $b=-an^2$. Case I. $n=0$ The equation becomes $$a(mx)^4+b(mx)^2=(ax^2+bx)(mx)^2$$Suppose that $x\ne0$. Dividing both sides by $(mx)^2$, $$a(mx)^2+b=ax^2+bx$$$$(am^2-a)x^2-bx+b=0$$This gives us $am^2-a=0\implies m=1,-1$ and $b=0$. Therefore, two possible family of solutions is given by $$P(x)=ax^2,~Q(x)=x~~~\text{where }a\in\mathbb R_{\ne0}$$$$P(x)=ax^2,~Q(x)=-x~~~\text{where }a\in\mathbb R_{\ne0}$$This is a special case of the solutions listed above. Case II. $b=-an^2$. The equation becomes $$a(mx)^4-an^2(mx)^2=(ax^2-an^2)(mx)^2$$Suppose that $x\ne0$. Dividing both sides by $a(mx)^2$ $$(mx)^2-n^2=x^2-n^2$$$$(m^2-1)x^2=0$$This shows us that $m=1,-1$. Therefore, another family of solutions is given by: $$P(x)=ax^2-an^2x,~Q(x)=x+n~~~\text{where }a\in\mathbb R_{\ne0}\text{ and }n\in\mathbb R$$$$P(x)=ax^2-an^2x,~Q(x)=-x+n~~~\text{where }a\in\mathbb R_{\ne0}\text{ and }n\in\mathbb R$$

$(P(x), Q(x)) = (ax^2, x), (ax^2, -x), (ax(x-4), x-2), (ax(x-4), 2-x)$ where $a\in \mathbb{R} \setminus \{0\}$.

Let the degree of $P$ and $Q$ be respectively $p$ and $q$. Then, the equation implies $2pq = p + 2q$ or $(p - 1)(2q - 1) = 1$, and thus $p = 2$ and $q = 1$. Second, obviously re-scaling the polynomial $P$ by multiplying it with non-zero constant does not affect the equation, so without loss of generality we can let it be monic. Also, letting the leading coefficient of $Q$ be $q_0$, we find that $(q_0^2)^2 = q_0^2$, so $|q_0| = 1$. Since $Q$ is a non-constant linear polynomial with real coefficients, there exists a real number $t$ such that $Q(t) = 0$. Since $Q$ is linear with leading coefficient $\pm 1$, either $Q(x) = x - t$ or $Q(x) = t - x$. Either way, $Q(x)^2 = (x - t)^2$. Plugging $x = t$ yields $P(0) = 0$, so $P(x) = xR(x)$ for some monic linear polynomial $R$. The original equation becomes $ (x - t)^2 R((x - t)^2) = xR(x) (x - t)^2 $, which simplifies to $$ R((x - t)^2) = xR(x) $$ Since $R(t^2) = 0$, we get $R(x) = x - t^2$. Plugging back yields $(x - t)^2 - t^2 = x(x - t^2)$ or $2t = t^2$, which implies $t = 0$ or $t = 2$.