If all zeros of $Q$ in $\mathbb C^2$ are zeros of $P$, then, by [Hilbert's Nullstellensatz][1], $Q$ 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][2].

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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.

  [1]: https://en.wikipedia.org/wiki/Hilbert%27s_Nullstellensatz
  [2]: https://mathoverflow.net/questions/468912/on-zeros-of-real-polynomials-in-two-variables/468918#comment1217727_468912