From $Q(2x) = Q(x)$, you deduce that $Q(2^x)$ and $Q(-2^x)$ are periodic in $x$ with period (dividing) $1$.
For a polynomial with any roots other than $x = 0$, this is a problem since it implies that the polynomial has infinitely many roots. From your deduced equation,
$$ (x−8)(2x−2)(2x−4)(2x−8)Q(2x)=8(x−1)(x−2)(x−4)(x−8)Q(x) \text{,} $$
cancelling common factors leaves $Q(2x) = Q(x)$ and you know $Q(x)$ is a polynomial. If $Q(x)$ has any roots (except at $x = 0$), it has infinitely many roots. The only polynomial with infinitely many roots is the polynomial $0$. Certainly, setting $Q(x) = 0$ makes the above equation true. Alternatively, let $Q$ be any constant other than $0$ and the above equation is satisfied without reducing it to the triviality "$0 = 0$". (We have two choices for $Q$: infinitely many roots or no roots, so either $Q = 0$ or $Q = c$ for any nonzero constant, $c$. You might wonder: what about $x^2 + 1$, which has no roots? It doesn't take any of its values infinitely many times. Remember every value $Q$ takes, it takes infinitely many times. Suppose $Q(1) = c$. Then $Q(x) - c$ is a polynomial with infinitely many roots, so it is the zero polynomial and we find $Q(x) = c$ for all $x$.)
Another way to see that $Q$ must be constant if $Q$ is not everywhere zero is this. Let $Q(1) = c$. Then $Q$ takes the value $c$ infinitely many times: at $\dots$, $1/8$, $1/4$, $1/2$, $1$, $2$, $4$, $8$, $\dots$. This means the polynomial $Q(x) -c$ has infinitely many zeroes, forcing $Q(x) - c = 0$, the zero polynomial. Then $Q(x) = c$ is constant.
We already know, since $Q(x)$ is constant, that $P(x)$ is cubic, so the proposed solutions are impossible. Neither of the answers you recite actually satisfy your functional equation.
\begin{align*}
(x-8)\left(3(2x)^6 - 27(2x)^4 + 33(2x)^2 - 1 \right) &= 192 x^7-1536 x^6-432 x^5+3456 x^4+132 x^3-1056 x^2-x+8 \text{,} \\
8(x-1)\left( 3x^6 - 27x^4 + 33 x - 1 \right) &= 24 x^7-24 x^6-216 x^5+216 x^4+264 x^3-264 x^2-8 x+8 \text{,} \\
(x-8)\left((2x)^6 - 33(2x)^4 + 27(2x)^2 - 3 \right) &= 64 x^7-512 x^6-528 x^5+4224 x^4+108 x^3-864 x^2-3 x+24 \text{, and} \\
8(x-1)\left(x^6 - 33x^4 + 27x^2 - 3 \right) &= 8 x^7-8 x^6-264 x^5+264 x^4+216 x^3-216 x^2-24 x+24 \text{.}
\end{align*}
It's easy enough to see that these don't even agree in their leading coefficients before multiplying them out: \begin{align*}
1 \cdot 3 \cdot 2^6 &\neq 8 \cdot 1 \cdot 3 \text{ and } \\
1 \cdot 2^6 &\neq 8 \cdot 1 \cdot 1
\end{align*}
