Let $1<m,n<p$, where $p$ is a prime number.
We want to prove that the polynomial $$f(X):=X^p-p^2X+p\in \Bbb Q[X]$$ has no roots in the field $E:=\Bbb Q(\sqrt[m]{n},\sqrt[n]{m})$.
My thought. Suppose on the contrary that $\alpha \in E$ is a root of this $f(X)$ inside $E$. Then, $f(\alpha)=0.$ Now, consider the extension of fields $$[E:\Bbb Q]=[E:\Bbb Q(\alpha)][\Bbb Q(\alpha):\Bbb Q].$$ By the Eisenstein's Criterion, $f$ is irreducible, monic and with root the element $\alpha$, so $m_{(\alpha,\Bbb Q)}=f$. Hence, $\deg m_{(\alpha,\Bbb Q)}=[\Bbb Q(\alpha):\Bbb Q]=p$ and therefore $p|[E:\Bbb Q]$. But now I m not sure that is in the correct direction.
Any help please?
You are almost there. Consider now $$[E:\Bbb Q]=[\Bbb Q(\sqrt[m]{n},\sqrt[n]{m}):\Bbb Q(\sqrt[m]{n})] \cdot [\Bbb Q(\sqrt[m]{n}) :\Bbb Q]$$
Since $p| [E:\Bbb Q]$ and $p$ is prime you get that $p|[\Bbb Q(\sqrt[m]{n},\sqrt[n]{m}):\Bbb Q(\sqrt[m]{n})]$ or $p| [\Bbb Q(\sqrt[m]{n}) :\Bbb Q]$.
Argue now that $$[\Bbb Q(\sqrt[m]{n},\sqrt[n]{m}):\Bbb Q(\sqrt[m]{n})] \leq n<p \\ [\Bbb Q(\sqrt[m]{n}) :\Bbb Q] \leq m <p$$