Let $\Phi_n$ be the $n$th cyclotomic polynomial: $${\Phi _{n}(x)=\!\!\prod _{\substack {1\leq k\leq n \\ \gcd(k,n)=1}} \!\!\big(x-e^{2i\pi {k/n}}\big).}$$

Here is a list of the first 30 cyclotomic polynomials.

Given an integer $k$, is there were always infinitely many integers $n$ such that $\Phi_n$ has $k$ terms, and if yes is it possible to classify them?

A friend of mine thinks there are infinitely many such $n$ for every prime $k$. However, $\Phi_{105}$ has $33$ terms so this does not happen just for prime $k$.

For $k=2$ there are infinitely many $n$ such that $\Phi_n$ has $k=2$ terms and I was told that this happens if and only if $n$ is a power of $2$. This is apparent in wikipedia's list, as the only cyclotomic polynomials with $2$ terms are $\Phi_1,\Phi_2,\Phi_4,\Phi_8$ and $\Phi_{16}$.

Taking $k=3$, the orders of the cyclotomic polynomials with $3$ terms does not match the powers of $3$: $\Phi_{12}=x^4-x^2+1$ and $\Phi_{24}=x^8-x^4+1$ both have $3$ terms and their orders, $12$ and $24$, are not powers of $3$ (though they are powers of $2$ times $3$, and the other orders $\leq 30$ of the cyclotomic polynomials with $3$ terms are precisely the powers of $3$).

For $k=5$, the only cyclotomic polynomials of order $\leq 30$ with $5$ terms are once again those whose orders are powers of $2$ times $5$. This does not extend to $k=7$ however as $\Phi_{15}$ has $7$ terms but its order isn't even disible by $7$.