Let $K=\mathbb F_{2^m}$ be a finite extension field over $L=\mathbb F_2$.
If every irreducible degree $m$ polynomial with coefficients in $L$ has a root $\alpha \in K$ such that the multiplicative group $ \{\alpha^n : n\ge 0\}$ is isomorphic to the multiplicative group $\mathbb F_{2^m}^{\times}$ (which is cyclic) , then is it true that $m$ is prime ? If this is true, then can we moreover say that $2^m-1$ is prime ?
The condition entails that (by action of the Galois group for instance, or uniqueness of $\mathbb{F}_{2^m}$) for all irreducible polynomials of degree $m$ and all roots of them in $K$, they are generators of the multiplicative group.
The condition implies that $2^m-1$ satisfies the following property: if $d|2^m-1$ and $d \neq 2^m-1$, then $d|2^u-1$ for some $u|m$.
Indeed, let $d|2^m-1$, let $\alpha \in \mathbb{F}_{2^m}^{\times}$ have order $d$. Let $F$ be the subfield of $K$ generated by $\alpha$ and let $\delta$ be the degree of its minimal polynomial. If $\delta=m$, we get a contradiction, because then $\alpha$ would generate the multiplicative group. So $\delta < m$ and hence $d|2^{\delta}-1$.
Now, the condition implies by Zsigmondy’s theorem ( see https://en.m.wikipedia.org/wiki/Zsigmondy's_theorem ) that either $2^m-1$ (hence $m$) is prime, or $m=6$. I leave you to check by hand the case $m=6$.
Note that by the same argument, the phenomenon cannot occur in any other base finite field (except maybe in cardinality $4,8,64$).