Number of prime ideals of a ring

The prime ideals of $\mathbb Q[x]/(x^m-1)$ correspond to prime ideals in $\mathbb Q[x]$ containing $x^m-1$, by taking the preimage under the surjection $\mathbb Q[x] \to \mathbb Q[x]/(x^m-1)$. Since $\mathbb Q[x]$ is a PID, those ideals are $(f)$ for $f$ an irreducible factor of $x^m-1$. We have the factorization $$x^m-1 = \prod_{d|m} \Phi_d$$ where $\Phi_d$ is the $d$th cyclotomic polynomial, which is irreducible. So the number of irreducible factors of $x^m-1$ is $\tau(m)$, the number of positive divisors of $m$.

A small addendum to marlu's answer: The Chinese Remainder Theorem implies $$\mathbb{Q}[x]/(x^m-1) \cong \prod_{d|m} \mathbb{Q}(\zeta_d),$$which is a finite direct product of fields. So the (prime) ideal structure is quite easy.