In the text by Dummit and Foote, there is the following exercise in chapter 13:
Determine the degree of the extension $\mathbb{Q}\big(\sqrt{3 + 2 \sqrt{2}}\big)$ over $\mathbb{Q}$.
To solve this problem, I proceeded by letting $\alpha = \sqrt{3 + 2 \sqrt{2}}$ and then computing $\alpha^2$ and $\alpha^4$. Assuming I did this correctly, one gets \begin{equation*} \alpha^2 = 3 + 2 \sqrt{2} \quad \text { and } \quad \alpha^4 = 17 + 12 \sqrt{2} \end{equation*}
It follows from this that the polynomial $f(x) = x^4 - 6x^2 + 1$ has $\alpha$ as a root. The only divisors of the constant term are $1$ and $-1$, and $f(\pm1) \neq 0$. By the Rational Roots Test, we should have $f$ be irreducible over $\mathbb{Q}$, so that the answer to the question is $4$.
However, I was looking at this write up of solutions, and here the answer is given as $2$. So I am guessing there is some error in my "proof" that the degree is $4$, but can't seem to find what I did incorrectly...
The trick is to note that $$ (1+\sqrt{2})^2=1+2\sqrt{2}+2=3+2\sqrt{2}$$ so in fact $\mathbb{Q}(\sqrt{3+2\sqrt{2}})=\mathbb{Q}(\sqrt{2})$.
The problem with your argument is that even if a quartic polynomial has no rational roots, it can still be reducible over $\mathbb{Q}$ since it could be a product of two irreducible quadratic polynomials.