Let $Q(x+3)=P(x)$. Hence $Q^2(x+3)-1=4Q(x^2-4x+4)=4Q((x-2)^2)$
Let $x\rightarrow x+2$. $Q^2(x+5)=4Q(x^2)+1$
So $Q(5+x)=\pm Q(5-x)$
Let $R(x)=Q(5+x)$. $R$ is either odd or even and $R^2(x)=4R(x^2-5)+1$
If $R$ is odd then $R(0)=0$ and $R(5)=\frac{1}{4}$ and $R(-20)=\frac{15}{64}$ and more generally :
$S_0=0$, $S_{n+1}=5-S_n^2$, $T_0=0$ and $T_{n+1}=\frac{1-T_n^2}{4}$
It's easy to see that $P(S_n)=T_n$, but $\lim(T_n)$ is finite and $\lim(|S_n|)=\infty$. So $R$ is not a polynomial function.
If $R$ is even, $R(0)$ is an optimum (either maximum or minimum, $R'(0)=0$). Hence $R(\sqrt{5})$ is an optimum too, and $S_0=0$, $S_{n+1}=\sqrt{5+S_n}$ , then $S_n$ are all optimum... But all $S_n$ are different, so $R'$ has an infinite number of roots. This is not a polynomial function, except for a degree 0 polynomial.
So the only solutions are constant.
I hope there is no mistake :)