Minimal polynomial of $\sqrt{3+2\sqrt{3}}$

You are unquestionably right. A much more advanced viewpoint:

The ring $R=\Bbb Z[\sqrt3\,]$ is the integer-ring of the quadratic number field $K=\Bbb Q(\sqrt3\,)$. It’s “well known” (and easy to prove) that $R$ is a Principal Ideal Domain (class number is $1$).

We have the factorization $z=3+2\sqrt3=\sqrt3(2+\sqrt3\,)$, in which the two factors are respectively, an indecomposable element, and a generator of the (free part of the) unit group. In particular, neither is a square, and they are independent modulo squares. Therefore, the square root $\sqrt z$ generates a quadratic extension of $K$. That does it.