Find the minimal extension field of $\mathbb{F}_2$ such that this extension contains an element of order $21$?
Attempt: I know that such an extension of $\mathbb{F}_2$ is like $\mathbb{F}_{2^s}$ and $2|s$. Such a field has a primitive element, say $\alpha$ that generated the whole field. We know by theory that such a primitive element is such that $\alpha^i =1 <=> 2^s-1|i$
So, $\alpha^{21}=1 <=> 2^s- 1 |21$
So I need to find the minimum $s$ such that $2^s - 1$ divides $21$. $s=3$ is the good candidate ($s=1$ corresponds to $\mathbb{F}_2$ which is the base field).
Therefore, such an extension is $\mathbb{F}_{2^2}=\mathbb{F}_4$
Is it correct?
No, it's incorrect. You need an element $a$ such that $a^{21}=1$, but $a^k\ne1$ when $0<k<21$.
Since the multiplicative group of a finite field is cyclic, you need to find the least exponent $m$ such that $21\mid(2^m-1)$, which is the reverse of what you're doing.
The group must have order divisible by $21$, and this suffices because the group is cyclic (actually abelian would suffice).
You therefore need $3\mid(2^m-1)$ and $7\mid(2^m-1)$. The former condition yields $m$ even, the latter that $3\mid m$.