why $\mathbb Z[\sqrt 2] \ncong \mathbb Z[\sqrt 3]$?

The following is a question from section $3.11$ of the book An introduction to abstract algebra by Allenby:

Explain intuitively why $\mathbb Z[\sqrt 2] \ncong \mathbb Z[\sqrt 3]$.back your intuition with proof.

Note:this example not only says that $\theta: a+b\sqrt 2 \mapsto a+b\sqrt 3$ is not isomorphism .It says no isomorphism can be found at all - no matter how clever choice of mapping you try to make ..

I can't see what's the intution behind this ..can anyone provide some hint on this...

Hint: $2$ is a square in the first ring. Is it a square in the second?

let $\sqrt {2}\in \mathbb {Z}[\sqrt{2}]$

$\phi(\sqrt {2 })^2= \phi (2) =2 $

$\phi (\sqrt {2 })= \pm \sqrt {2} \not\in \mathbb {Z} \sqrt{3} $