Assuming $x^2 + x + [1]$ is reducible in $\mathbb{Z}_2 [x]$
$$x^2 + x + [1] = (x + [a])(x + [b]) = x^2 + [a]x + [b]x + [a][b] = x^2 + ([a]+[b])x + [a][b]$$
for some $[a],[b] \in \mathbb{Z}_2$
This means that
$$[a]+[b] = [a+b] = [1] \therefore[a]=[1]-[b]=[1-b]$$ $$[a][b]=[ab]=[1] \therefore [a]=[b]^{-1}$$
Putting these two equations together,
$$[b-1] = [b]^{-1}$$
Inspecting the elements of $\mathbb{Z}_2$
$$[1] = [-1] = [0-1] \neq [0]^{-1}=[0]$$ $$[0]=[1-1] \neq [1]^{-1}=[1]$$
$\square$