Q[\sqrt{2}] not isomorphic to Q[\sqrt{3}]

Show that $\mathbb{Q}[\sqrt{2}] \not\cong \mathbb{Q}[\sqrt{3}]$

$\mathbb{Q}[\sqrt{2}]=\{a+b\sqrt{2}|a,b\in{\mathbb{Q}}\}$ and $\mathbb{Q}[\sqrt{3}]= \{c+d\sqrt{3}|c,d\in{\mathbb{Q}}\}$.

$\textit{Proof}$: Assume to the contrary that $\mathbb{Q}[\sqrt{2}] \cong \mathbb{Q}[\sqrt{3}]$. So, there exists an isomorphism between these two fields. First, we justify but do not prove that there is no problem of identification with elements between these two fields. In other words the number say 2 in $\mathbb{Q}[\sqrt{2}]$ is the same as the number 2 in $\mathbb{Q}[\sqrt{3}]$. With an isomorphism between them, $$\varphi(1)=1$$ $$\varphi(1+1)=\varphi(1)+\varphi(1)=1+1=2$$ And so on, so the integers must be fixed by this isomorphism. With the integers fixed, then the field of fraction of the integers, which is $\mathbb{Q}$ must be fixed. In other words, the ground field of these two field extensions must be fixed. This is consistent with the idea in Galois theory that automorphisms of field extensions fix the ground field.
Isomorphisms fix structures. In $\mathbb{Q}[\sqrt{2}]$ there is an element, namely $\sqrt{2}$ such that when squared, we get the number 2. Therefore, there must exist an element in $\mathbb{Q}[\sqrt{3}]$ such that when squared, we get 3.
$(c+d\sqrt{3})^2=2$
$c^2+2cd\sqrt{3}+3d^2=2$. $(\star)$
$c,d\in{\mathbb{Q}}$ and $cd\sqrt{3} \not \in{\mathbb{Q}}$, so $cd\sqrt{3}=0 \Longrightarrow c=0$ or $d=0$ since we are in a field.
If $c=0$ then $(\star)$ becomes
$3d^2=2 \Longrightarrow d=\pm \sqrt{\frac{2}{3}} \not \in {\mathbb{Q}}$...a contradiction
Id $d=0$, then $(\star)$ becomes
$c^2=2 \Longrightarrow c=\pm \sqrt{2} \not \in{\mathbb{Q}}$...a contradiction.
In any case, an isomorphism between $\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q}[\sqrt{3}]$ leads to a contradiction.
$\therefore \quad \mathbb{Q}[\sqrt{2}] \not\cong \mathbb{Q}[\sqrt{3}]$