Is $\mathbb{Q}[\sqrt{n}]$ ever isomorphic to $\mathbb{Q}[\sqrt{m}]$ where m,n are not perfect squares?

Question

I was doing some abstract algebra exercises and I noticed that we if we assume there exists some homomorphism $\phi : \mathbb{Q}[\sqrt{n}] \rightarrow \mathbb{Q}[\sqrt{m}]$. Then if we want an isomorphism then $\phi$ must be surjective which means that $\phi(1) = 1$. Which means that $\phi (a) = a $ for some $a\in$ $\mathbb{Q}$. Then for some $(a + b\sqrt{n})$ we get that $$\phi(a + b\sqrt{n}) = a +b\phi(\sqrt{n})$$ Now we also have that, $$\phi (n) = n = \phi ((\sqrt{n})^2) = (\phi(\sqrt{n}))^2$$ Then if $\phi(\sqrt{n}) = c + d\sqrt{m}$ we get that $$n = (\phi(\sqrt{n}))^2 = (c + d\sqrt{m})^2$$ $$ n = c^2 + 2\sqrt{m}cd + md^2$$ we get that $cd = 0$ so we have that if $c = 0 $ $$ n = md^2 \Rightarrow d = \sqrt{\dfrac{n}{m}}$$ which is a contradiction as $d\in \mathbb{Q}$ similarly we get that if $ d = 0$ $$ n = c^2 \Rightarrow c = \sqrt{n} $$ which again contradicts that $c\in \mathbb{Q}$ which means that we cannot construct an isomorphism. Is there something that I am assuming that I shouldn't be or is this correct?