Let $c,d \in Z$ that aren't the square other number. Then we have to prove that $Q(\sqrt{c}) = Q(\sqrt{d})$ iff $cd$ is a squared number.
What I tried to do is as follows:
Since both $c$ and $d$ aren't squares, it means that their roots aren't rational. Therefore, their irreductible polynomials are $p(x) = x^2 - c$ and $q(x) = x^2 - d$. If we suppose that $Q(\sqrt{c}) = Q(\sqrt{d})$ then $\sqrt{d} = a + b\sqrt{c}$ since $\{1, \sqrt{c}\}$ is a Q-base of $Q(\sqrt{d})$. Y expected to get some result playing with that expression, but I didn't seem to find anything useful, and I can't think of other approach.
So, following @PhysMath answer and using @TokenToucan suggestion, that means that $2ab\sqrt{c}$ should be zero, so either $a$ or $b$ must be zero.
Now, $b$ can't be zero since if that was the case, $\sqrt{d} = a$ which cannot happen. Therefore $a = 0$ and $d = b^2c$, so finally $cd = b^2c^2 = (bc)^2$.
For the other implication following the steps in reverse should work, since that way you can find the original identity.
I hope I'm right.