Show that $I=\langle x+1, x^2+1\rangle$ is maximal in $\mathbb Z[x]$. I’m trying to show this by showing that $\mathbb Z[x]/I$ is a field. Since in this ring $x=-1$ and so $x^2=1$, but also $x^2=-1$ from which it follows that $1=-1$ and so it is probably isomorphic to $\mathbb F_2$? But I can’t find a homomorphism from $\mathbb Z[x]\ \longrightarrow\ \mathbb F_2$ with kernel $I$.

Very useful exercise: If $R$ is a commutative ring and $r_1,\ldots,r_n\in R$, then
$$R/\langle r_1,\ldots,r_n\rangle\cong (R/\langle r_1\rangle)/\langle\bar{r_2},\ldots,\bar{r_n}\rangle.$$
Applying this here makes the exercise very easy; we have $R=\Bbb{Z}[x]$ and $r_1=x+1$ and $r_2=x^2+1$. Then
$$\Bbb{Z}[x]/\langle x+1,x^2+1\rangle\cong(\Bbb{Z}[x]/\langle x+1\rangle)/\langle\overline{x^2+1}\rangle.$$
Of course $\Bbb{Z}[x]/\langle x+1\rangle\cong\Bbb{Z}$ by mapping $x$ to $-1$. Then $x^2+1$ is mapped to $(-1)^2+1=2$ and so
$$(\Bbb{Z}[x]/\langle x+1\rangle)/\langle\overline{x^2+1}\rangle\cong\Bbb{Z}/\langle2\rangle=\Bbb{F}_2.$$
This is a field, and so this shows that the original ideal is maximal.

Alternatively, you mention that you already suspect that the quotient is isomorphic to $\Bbb{F}_2$, but cannot find a homomorphism $\Bbb{Z}[x]\ \longrightarrow\ \Bbb{F}_2$ with kernel $I$. Note that such a homomorphism is determined entirely by where $x$ is mapped. So $x$ must map to some element of $\Bbb{F}_2$ such that $x+1$ and $x^2+1$ are mapped to $0$. There aren't many candidates; you just have to check that this does indeed work.
