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 $n$.