The OP has already suggested two ways to solve this:
(a) Let $\zeta_n$ generated the group of roots of unity in $\mathbb Q(\zeta_l)$. Then $2l$ divides $n$, and
also $\mathbb Q(\zeta_l) = \mathbb Q(\zeta_n)$. A consideration of degrees shows that $\varphi(n) = \varphi(l)$, and combining this with the fact that $2l$ divides $n$, elementary number theory implies that in fact $n = 2l$.
(b) Ramification theory rules out the possibility of $\zeta_n$ lying in $\mathbb Q(\zeta_l)$ if $n$ is divisible by an odd prime $p \neq l$ or by a power of $2$ greater than the first.
Here are some other arguments (I continue to let $\zeta_n$ be the generator of the roots of unity in $\mathbb Q(\zeta_l)$):
(c) Galois theoretic: since $\mathbb Q(\zeta_l) = \mathbb Q(\zeta_n)$,
passing to Galois groups over $\mathbb Q$, we find that the units in $\mathbb Z/n$ project isomorphically onto the units in $\mathbb Z/l$. Given that $2l | n$, we deduce from the Chinese remainder theorem that $n = 2l$.
(d) Discriminants: Since $\mathbb Q(\zeta_l) = \mathbb Q(\zeta_n)$, a consideration of the standard discriminant formulas shows that $n = 2l$.
(e) Looking at the reduction modulo split primes: Choose $p$ prime to $n$ and congruent to $1$ mod $l$. Then the group of $n$th roots of unity injects
into the residue field of any prime lying over $p$. Since $p \equiv 1 \bmod l$, this residue field is just $\mathbb F_p$, and so we find that $n | p-1$
if $p > n$ (say) and $p \equiv 1 \bmod l$. Dirichlet's theorem then gives that the units in $\mathbb Z/n$ project isomorphically onto the units in $\mathbb Z/l$, from which we deduce that $n = 2l$.
(f) Working locally at l: it is not hard to check that the roots of unity in $\mathbb Q_l(\zeta_l)$ are precisely $\mu_{l(l-1)}$. So we have to show that the only $(l-1)$st roots of $1$ in $\mathbb Q(\zeta_l)$ are $\pm 1$. Actually I don't see how to do this right now without reverting to one of the other arguments, but there's probably a pithy way.
Note that (c) is just a fancy version of (a), while (d) is a more concrete form of (b) (which uses less theory). It may seem that (e) is overkill, and it certainly is for this question, but the method can be useful, and it has an obvious connection to (c) via reciprocity laws. Method (f) (unfortunately incomplete) is related to (b).
The degree of $e^{2\pi i/n}$ goes to infinity with $n$. If $K$ had an infinity of roots of unity, it would have elements of arbitrarily high degree, and thus would not be of finite degree over the rationals, and thus would not, in fact, be an algebraic number field.
Best Answer
Let $w$ be the number of roots of unity in $K$, so $w$ is even. If $p$ is a prime factor of $w$ then $\mathbf Q(\zeta_p) \subset K$. Since $(p) = (1-\zeta_p)^{p-1}$ in $\mathbf Z[\zeta_p]$, every prime $\mathfrak p$ over $p$ in $K$ ramifies when $p > 2$. Hence when a prime $\mathfrak p$ in $K$ is unramified and $\mathfrak p$ doesn't lie over $2$, $\mathfrak p \nmid (w)$. Then $x^w - 1 \bmod \mathfrak p$ is separable and splits completely, so $w \mid ({\rm N}(\mathfrak p)-1)$.
There is a converse result: if $d \mid ({\rm N}(\mathfrak p)-1)$ for all but finitely many unramified $\mathfrak p$, then $d \mid w$, so $w$ is the gcd of the integers ${\rm N}(\mathfrak p)-1$ as $\mathfrak p$ runs over any set of all but finitely many unramified primes in $K$. This may not look like a practical way of computing $w$, but if you can compute norms of prime ideals in $K$ then you can look at a large finite set of such numbers ${\rm N}(\mathfrak p) - 1$ to make a plausible guess at the gcd of all such numbers (with $\mathfrak p$ unramified and not lying over $2$) in order to guess a value for $w$.
Example. Take $K = \mathbf Q(i)$, for which $w = 4$. When $(\pi)$ is a prime in $\mathbf Q(i)$ other than $(1+i)$ then ${\rm N}(\pi)-1$ is divisible by $4$, either by a direct calculation (since ${\rm N}(\pi)$ is a prime that's $1 \bmod 4$ or is $p^2$ where $p \equiv 3 \bmod 4$) or because $\mathbf Z[i]/(\pi)$ is a field containing a primitive $4$th root of unity. If $(\pi)$ runs over all but finitely many primes in $\mathbf Q(i)$ other than $(1+i)$ then it can be proved that the numbers ${\rm N}(\pi)-1$ have gcd $4$. This generalizes the fact that if $m \geq 2$ then and $p$ runs over all but finitely many primes that are $1 \bmod m$, then the numbers $p-1$ have gcd $m$ (proved by Dirichlet's theorem).