As often happens when dealing with $\mathbf{Z}_n$, the Chinese remainder theorem is your friend. If the prime factorization of $n$ is
$$
n=\prod_i p_i^{a_i},
$$
then by CRT we have an isomorphism of rings
$$
\mathbf{Z}_n\cong\bigoplus_i \mathbf{Z}_{p_i^{a_i}}.
$$
Observe that the isomorphism maps the residue class of an integer $m$ (modulo $n$) to a vector with all the components equal to the residue class of $m$ (this time modulo various prime powers):
$$
\overline{m}\mapsto(\overline{m},\overline{m},\ldots,\overline{m}).
$$
So the residue class of $m$ is an idempotent if and only if it is an idempotent modulo all the
prime powers $p_i^{a_i}$.
Let us look at the case of a prime power modulus $p^t$. The congruence $x^2\equiv x\pmod{p^t}$ holds, iff $p^t$ divides $x^2-x=x(x-1)$.
Here only one of the factors of, $x$ or $(x-1)$, can be divisible by $p$, so for the
product to be divisible by $p^t$ the said factor then has to be divisible by $p^t$. Thus we can conclude that $x\equiv 0,1 \pmod{p^t}$ are the only idempotents modulo $p^t$.
Therefore we require that
$$
m\equiv 0,1\pmod{p_i^{a_i}}
$$
for all $i$. By CRT these congruences are independent for different $i$, so the number of pairwise non-congruent idempotents is equal to $2^\ell$, where $\ell$ is the number of distinct prime factors $p_i$ of $n$.