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).