Show that there are infinitely many reducible polynomials of the form $x^n+x+1$ in $\mathbf{F}_2[x]$

As a variation on the (essentially equivalent) ideas in the answers and comments: we could ask that there be $\alpha\in \mathbb F_4$ solving $x^n+x+1=0$, noting that certainly there is no solution in $\mathbb F_2$. For $\alpha\in \mathbb F_4$, $\alpha^4=\alpha$, and since $\alpha\not=0,1$, also $\alpha^3=1$ and $\alpha^2+\alpha+1=0$. Thus, $$ \alpha^{3n+2} + \alpha + 1 = \alpha^2+\alpha+1 = 0 $$ Thus, $x^2+x+1$ divides $x^{3n+2}+x+1$.

Proof by induction.

Base case ($n=0$): $$\begin{align} (x^2+x+1)\left(x^3+\sum\limits_{i=0}^0 (x^{3i}+x^{3i+2})\right)&=(x^2+x+1)(x^3+x^2+1)\\&=x^5+2x^4+2x^3+2x^2+x+1\\&=x^5+x+1=x^{3(0)+5}+x+1 \end{align}$$ (working in $\mathbb{F}_2[x]$).

Inductive hypothesis ($n \geq 0$): Suppose that $$(x^2+x+1)\left(x^{3(n+1)}+\sum\limits_{i=0}^n (x^{3i}+x^{3i+2})\right)=x^{3n+5}+x+1$$

Then: $$\begin{align} &(x^2+x+1)\left(x^{3(n+2)}+\sum\limits_{i=0}^{n+1} (x^{3i}+x^{3i+2})\right)=\\ &(x^2+x+1)\left(x^{3(n+2)}+x^{3(n+1)}+x^{3(n+1)+2}+\sum\limits_{i=0}^{n} (x^{3i}+x^{3i+2})\right)=\\ &(x^2+x+1)(x^{3(n+2)}+x^{3(n+1)+2}) + (x^2+x+1)\left(x^{3(n+1)}+\sum\limits_{i=0}^{n} (x^{3i}+x^{3i+2})\right) \end{align}$$

using our inductive hypothesis we get

$$\begin{align} &=(x^2+x+1)(x^{3(n+2)}+x^{3(n+1)+2}) + x^{3n+5}+x+1\\ &=(x^2+x+1)(x^{3n+6}+x^{3n+5}) + x^{3n+5}+x+1\\ &=x^{3n+8}+2x^{3n+7}+2x^{3n+6}+2x^{3n+5}+x+1\\ &=x^{3(n+1)+5}+x+1 \end{align}$$

Therefore, $x^{3(n+1)+5}+x+1$ is reducible in $\mathbb{F}_2[x]$ (or in any polynomial ring with coefficients in a field of characteristic 2) for all non-negative integers $n$.