Minimal polynomial of $\sqrt{2+\sqrt[3]{3}}$ over $\mathbb{Q}$

Here is an alternative solution. Working in $\mathbb F_3$, the minimal polynomial for $\alpha$ factorises as $(x^2+1)^3$, and $x^2+1$ is irreducible. Thus 2 divides $[\mathbb Q(\alpha):\mathbb Q]$, and you showed 3 divides this as well.

This is not an alternative solution, but I think I can continue from where you have left.

So you were trying to show that there is no rational solution to $$ 2 + \sqrt[3]{3} = (a + b \sqrt[3]{3} + c\sqrt[3]{9})^2 $$

Taking the field norm (in $\mathbb{Q}(\sqrt[3]{3})$ over $\mathbb{Q}$) of both sides, you get $11 = r^2$ for some $r\in\mathbb{Q}$ which is absurd.