On irreducible factors of $x^{2^n}+x+1$ in $\mathbb Z_2[x]$

Let $p(x)\mid x^{2^n}+x+1$, $p$ irreducible, and $a\in\overline{\Bbb F}_2$ a root of $f$. Then $a^{2^n}+a+1=0$, equivalently $a^{2^n}=a+1$. It follows that $a^{2^{2n}}=a$, so $\Bbb F_2(a)\subset\Bbb F_{2^{2n}}$. This shows that $\deg p\mid 2n$.

An answer along the idea you had is that $t(x) = x^{2^n} + x$ is the function that computes the trace of the quadratic extension $\mathbb{F}_{2^{2n}} / \mathbb{F}_{2^n}$. Thus every element of $\mathbb{F}_{2^{2n}}$ of trace $1$ over $\mathbb{F}_{2^n}$ is a root of $f(x)$. As there are $2^n$ of these, they are all of the roots of $f(x)$.