If all zeros of $Q$ in $\mathbb C^2$ are zeros of $P$, then, by Hilbert's Nullstellensatz, $Q$ divides $P^r$ for some natural $r$.
If $Q$ is also irreducible, then it follows that $Q$ divides $P$.
If $Q$ is not irreducible, then $Q$ does not have to divide $P$, as noted in my comment.
Now, how to answer the original question, for real (rather than complex) zeros -- assuming $Q$ is irreducible and the set of real zeros of $Q$ is of cardinality $>1$?
Using an affine transformation, without loss of generality $Q(x,y)$ is $x^2+y^2-1$ or $x^2-y^2-1$ or $y-x^2$.
If $Q(x,y)=x^2+y^2-1$, then $p_1(t):=P(\cos t,\sin t)=0$ for all real $t$ and hence for all complex $t$, since $p_1$ is an entire function. So, in this case, all complex zeros of $Q$ are zeros of $P$.
If $Q(x,y)=x^2-y^2-1$, then $p_2(t):=P(\cosh t,\sinh t)=0$ for all real $t$ and hence for all complex $t$, since $p_2$ is an entire function. So, all complex zeros of $Q$ are zeros of $P$ in this case as well.
If finally $Q(x,y)=y-x^2$, then $p_3(t):=P(t,t^2)=0$ for all real $t$ and hence for all complex $t$, since $p_3$ is an entire function. So, all complex zeros of $Q$ are zeros of $P$ in this latter case as well.
So, it remains to use the first two sentences of this proof.
A more general result is now available.