Let $f(T)=T^3+T^2+2T+2 \in \mathbb{Z}[T]$ and let $I$ be the ideal of the ring $\mathbb{Z}[T]$ generated by $f(T)$ and $5$. Find the number of elements $z \in \mathbb{Z}[T]/I$ such that $z^2=z$.
Some observations:
$\mathbb{Z}[T]/I \cong \{ aT^2 + bT + c \, | \, a,b,c \in \mathbb{Z}_5 \}$.
Now clearly $0$ and $1$ are solutions. Actually for every $z$, $c \in \{ 0, 1\}$
Beside $0$ and $1$, there is no solution such that $a \neq 0$
Now I wanted to test all the elements satisfying 2 and 3, but I am wondering if there is a faster solution, because the next exercise is to find all $z$ such that $z^{18} = 1$.
Since your polynomial is reducible, write it as $T^3 + T^2 + 2T + 2 = (T + 1)(T^2 + 2)$. Note that these factors are also irreducible modulo $5$ so
$$\Bbb Z[T]/(5, f(T)) \cong \Bbb F_5[T]/(\overline{f}(T)) \cong \Bbb F_5 \times \Bbb F_{25}$$
by the Chinese remainder theorem.
Since this ring is finite, it is Artinian, so there are $2^m$ idempotent elements where $m$ is the number of maximal ideals of the ring. In a direct product $R \times S$ of rings, the only maximal ideals are those of the form $R \times \mathfrak{m}_S$ or $\mathfrak{m}_R \times S$.
In this case, $R = \Bbb F_5$ and $S = \Bbb F_{25}$, both fields. There are very few maximal ideals in fields, so this should give you what you need to count the maximal ideals of the product.