Suppose for a contradiction that $x^2-7$ is reducible over $\mathbb{Q}(\sqrt[5]{3})$. Then $\sqrt{7}\in\mathbb{Q}(\sqrt[5]{3})$. It follows that $\mathbb{Q}\subset\mathbb{Q}(\sqrt{7})\subset\mathbb{Q}(\sqrt[5]{3})$. Because $x^2-7$ is the minimal polynomial of $\sqrt{7}$ over $\mathbb{Q}$, $[\mathbb{Q}(\sqrt{7}):\mathbb{Q}]=\deg(x^2-7)=2$.
Consider now the polynomial $q(x)=x^5-3$ which has the root $\sqrt[5]{3}$. It is in fact the minimal polynomial of $\sqrt[5]{3}$ over $\mathbb{Q}$ by Eisenstein's criterion but we continue without this knowledge. We know that $q$ has no rational roots by the rational root test, $q$ can therefore not have any linear factors. Since the minimal polynomial of $\sqrt[5]{3}$ divides $x^5-3$, it must therefore be of degree $2$, $3$ or $5$.
However note also that $[\mathbb{Q}(\sqrt[5]{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt[5]{3}):\mathbb{Q}(\sqrt{7})][\mathbb{Q}(\sqrt{7}):\mathbb{Q}]=[\mathbb{Q}(\sqrt[5]{3}):\mathbb{Q}(\sqrt{7})]\cdot2$ which implies that the degree of the minimal polynomial of $\sqrt[5]{3}$ must be divisible by $2$. The polynomials degree must thus equal $2$. In that case $2=[\mathbb{Q}(\sqrt[5]{3}):\mathbb{Q}(\sqrt{7})]\cdot2$ which implies that $\mathbb{Q}(\sqrt[5]{3})=\mathbb{Q}(\sqrt{7})$.
My question is then, is there a way to continue without using Eisenstein's criterion mentioned above? Is there some more elementary way of showing $\mathbb{Q}(\sqrt[5]{3})\neq\mathbb{Q}(\sqrt{7})$ which would then in turn prove the irreducibility of $x^2-7$ over $\mathbb{Q}(\sqrt[5]{3})$?
