I have a slight confusion about this argument below. Why did we need to show that there is no $\alpha\in \mathbb{F}_{2^2}^{\times}$ instead of $\mathbb{F}_{2}^{\times}$ in order to conclude?
We use the 5th cyclotomic polynomial $\Phi_5(x)=\frac{x^5-1}{x-1}=x^4+x^3+x^2+x+1$. Let $\mu_5$ denote the 5th roots of unity. It is enough to show that there are no linear factors or degree 2 factors. So we start by showing that there are no linear factors.
See that if there is an $\alpha\in \mathbb{F}_2$ such that $\alpha^5=1, \alpha^n\neq 1$ for $1\leq n\leq 4$, then $\alpha$ is a root of $\Phi_5(x)$. We also note that $\alpha$ is a unit in $\mathbb{F}_2$ because we can define $\alpha^{-1}=\alpha^4$. As a corollary of Lagrange's theorem, we know that for any $\alpha\in \mu_5$, $|\alpha|=5$ does not divide the size of the set of units $|\mathbb{F}_2^{\times}|=1$. This means $\alpha\notin \mathbb{F}_2^{\times}$ and there is no linear factor for $\Phi_5(x)$.
Now we do a similar argument to show that there is no quadratic factor.
If $\Phi_5(x)$ has a quadratic factor, then its root would come from $\mathbb{F}_{2^2}^{\times}$. (why is this?) But we notice that for every element $\alpha \in \mu_5$, $|\alpha|=5$ and $|\mathbb{F}_{2^2}^{\times}|=3$. $5$ doesn't divide $3$, which implies that no 5th root of unity is in $\mathbb{F}_{2^2}^{\times}$. Therefore, there are no quadratic factor either and $\Phi_5(x)$ is irreducible.