If all zeros of $Q$ 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].


  [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