Let $\zeta_n:=$ primitive $n$-th root of $1$ over $\mathbb Q$. And let $\alpha \in \mathbb Q[\zeta_n]$ satisfy $\alpha^m=2$ for some positive integer $m$. Then I have to show that $m=1$ or $2$.
We have the tower of fields $\mathbb Q \subset \mathbb Q(\alpha) \subset \mathbb Q(\zeta_n)$ with $|\mathbb Q(\zeta_n):\mathbb Q|=\phi(n)$ and $|\mathbb Q(\alpha):\mathbb Q|=m$. Since $Gal(\mathbb Q(\zeta_n)/\mathbb Q)$ is abelian $\mathbb Q(\alpha)/\mathbb Q$ is also Galois extension. Also clearly $m \mid \phi(n)$. How can I conclude that $m=1$ or $2$. I need some help. Thanks.
Based on discussion with Jyrki Lahtonen in comment I am writing down an answer.
Since $Gal(\mathbb Q(\zeta_n)/\mathbb Q) \cong (\mathbb Z/n\mathbb Z)^{*}$, $Gal(\mathbb Q(\zeta_n)/\mathbb Q)$ is abelian. Hence every intermediate extension of $\mathbb Q(\zeta_n)/\mathbb Q$ is Galois. As $\alpha \in \mathbb Q(\zeta_n)$, $\mathbb Q(\alpha)/\mathbb Q$ is Galois, and hence normal extension. Thus $\mathbb Q(\alpha)$ is the splitting field of $x^m-2$ over $\mathbb Q$. In particular, $\mathbb Q \subset \mathbb Q(\sqrt[m]{2})\subset Q(\alpha)$, where $|\mathbb Q(\alpha):\mathbb Q|=|\mathbb Q(\sqrt[m]{2}):\mathbb Q|=m$. Thus we have $\mathbb Q(\alpha)=\mathbb Q(\sqrt[m]{2})$. Hence $\alpha \in \mathbb R$. Now for a natural number $n$, all roots of $x^n-1$ are real if and only if $n\leq 2$. Since $\mathbb Q(\alpha)$ is a splitting field of $x^m-2$ over $\mathbb Q$, $m=1,2$ are the only possibilities.