Note that if $r$ is a root of $Q(x)$ or $P(x)$, it must also be a root of $Q(x^2)$, which says that $r^2$ is also a root of $Q(x)$. Moreover, of the $24$ square roots of the $12$ roots of $Q(x)$ (by multiplicity), half are the roots of $Q(x)$ and half are the roots of $P(x)$. This cuts down the possibilities considerably.
EDIT:
Consider a directed graph with vertices the $24$ roots (counted by multiplicity) of $P$ and $Q$; if root $r$ is not $0$ or $1$, so that $r^2 \ne r$, an arc goes from $r$ to (one of the instances of) $r^2$. $12$ of the vertices which have in-degree $0$ are coloured red (corresponding to roots of $P$), the rest are blue.
The only possible terminal vertices have values $0$ and $1$. $0$ is isolated, while $1$ has in-degree $1$, corresponding to $-1^2 = 1$. All other vertices have in-degree $2$, corresponding to their two square roots.
A blue vertex could be on a $k$-cycle, corresponding to a root $r \ne 0,1$ with $r^{2^k} = r$ (and $r^{2^j} \ne r$ for $0 < j < k$). The minimal polynomial of such a root must be a cyclotomic polynomial $\Phi_m(x)$
where $m$ divides $2^k-1$ (and not $2^j-1$ for $0 < j < k$), and this cyclotomic polynomial must divide $Q(x)$. Its degree is $\varphi(m)$ where $\varphi$ is the Euler totient function.
Roots $r$ that are not on a cycle but such that $r^2$ is in a cycle that corresponds to polynomial $\Phi_m(x)$ are roots of $\Phi_m(x^2)/\Phi_m(x)$.
Roots $r$ that are $k$ steps removed from a cycle are roots of $\Phi_m(x^{2^k})/\Phi_m(x^{2^{k-1}})$. I'm not sure that these are necessarily irreducible, so you might have only some of the factors in $Q(x)$ and the others in $P(x)$.
The odd integers $m$ with $\varphi(m) \le 12$ are
$1, 3, 5, 7, 9, 11, 13, 15, 21$, which have $\varphi(m) = 1, 2, 4, 6, 6, 10, 12, 8, 12$. It should be possible (but a bit tedious) to work out all possibilities.
