Polynomial irreducible - maximal ideal

In a commutative ring $R$ with $1$ and $a\ne 0$ \begin{array}\ R/(a) \text{ integral domain} &\iff &(a) \text{ prime ideal} & \iff & a \text{ prime}\\ &&& & & \\ \Uparrow&&(a) \text{ maximal among principal} & \Longleftarrow & a \text{ irreducible} &\\ && & & & \\ R/(a) \text{ is a field} & \iff &(a) \text{ maximal ideal} \end{array}

In an integral domain $R$ and $a\ne 0$ \begin{array}\ R/(a) \text{ integral domain} &\iff &(a) \text{ prime ideal} & \iff & a \text{ prime}\\ &&& & \Downarrow & \\ \Uparrow&&(a) \text{ maximal among principal} & \iff & a \text{ irreducible} &\\ && & & & \\ R/(a) \text{ is a field} & \iff &(a) \text{ maximal ideal} \end{array}

In a UFD $R$ and $a\ne 0$

\begin{array}\ R/(a) \text{ integral domain} &\iff &(a) \text{ prime ideal} & \iff & a \text{ prime}\\ &&& & \Updownarrow & \\ \Uparrow &&(a) \text{ maximal among principal} & \iff & a \text{ irreducible} &\\ && & & & \\ R/(a) \text{ is a field} & \iff &(a) \text{ maximal ideal} \end{array}

In a PID $R$ and $a\ne 0$ \begin{array}\ R/(a) \text{ integral domain} &\iff &(a) \text{ prime ideal} & \iff & a \text{ prime}\\ && & & \Updownarrow & \\ \Uparrow &&(a) \text{ maximal among principal} & \iff & a \text{ irreducible} &\\ &&\Downarrow & & & \\ R/(a) \text{ is a field} & \iff &(a) \text{ maximal ideal} \end{array}

There is a theorem that you can use: Given a commutative ring $R$ with identity, $I$ is a maximal ideal in $R$ if and only if $R/I$ is a field. (Similarly, $I$ is a prime ideal if and only if $R/I$ is an integral domain.)

What do elements in $\mathbb{Z}[x]/\langle x - 1\rangle$ look like? Do they form a field?

In 3, your argument is wrong. In an integral domain, $a$ is irreducible iff $(a)$ is maximal among principal ideals, but $\mathbb{Z}[x]$ is not a PID, thus you cannot conclude that $(a)$ is maximal. In fact it is not, because $\mathbb{Z}[x]/\langle x-1\rangle$ is not a field. So for example $\langle 2,x-1\rangle$ is a maximal ideal containing $\langle x-1\rangle$.