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
1). This is not so straightforward. As stated, your hypotheses hint at Kummer theory, and at a cohomological description of Brauer groups which you can find e.g. in Gille & Szamuely book "Central Simple Algebras and Galois Cohomology", chap. 4 (Cambridge Studies, 2017). Let us first recall the following general results : for any field $F$, denote by $F_{sep}$ a separable closure, and $G_F=Gal(F_{sep}/F)$. Then $Br(F)$ is canonically isomorphic to the Galois cohomology group $H^2(G_F, {F_{sep}}^*)$ . It is pointed out in G-S's book that this is much more tractable than the usual description $Br(F)\cong H^1(G_F,PGL_{\infty})$. For instance, given a Galois extension of degree $n$ and group $G$, Hilbert's thm. 90 (= nullity of $H^1(G, E^*))$ gives rise to an exact inflation-restriction sequence $0\to H^2(G, E^*) \to Br(F) \to Br(E)$, which means that $H^2(G, E^*)$ is canonically isomorphic to the relative Brauer group $Br(E/F)$ := the subgroup of classes of $Br(F)$ which split in $E$. It follows also easily Hilbert's thm. 90 that $Br(E)[n]\cong H^2(G_E, \mu_n)$, where $\mu_n$ is the group of $n$-th roots of unity.
2). It is classically known from cohomology theory that the order $n$ of $G$ kills $H^2(G, E^*)$, i.e. $Br(E/F)\subset Br(E)[n]$ (@). We would like to have a stronger hold on property (@), at least when $G$ is cyclic, in which case $H^2(G,M)\cong M/N(M)$ for any $G$-module $M$, where $N$ denotes the norm map of $E/F$. Under your hypotheses, since $G$ acts trivially on $\mu_n$, Kummer theory yields an isomorphism $\chi: G\cong Hom(G_F, \mu_n)=H^1(G_F,\mu_n)$, and property (@) reads: $\mu_n \subset F^*/N(E^*)$. Using basic properties of the cohomological cup-product, it's not hard to check that the isomorphism $F^*/N(E^*)\cong H^2(G,E^*)$ sends the class $a$ mod $N(E^*)$ to the class of the cyclic algebra $(\chi, a)$ (which is killed by the order $n$ of $G$). In particular, the class of $(\chi, a)$ splits in $E$ iff $b$ is a norm in the extension $E/F$. Note that Kummer theory allows to write $E=F(\sqrt [n] b)$, and the definition of $\chi$ then shows that $(\chi, a)$ splits iff $a$ is a norm from $F(\sqrt [n] b)$ (this statement is symmetrical w.r.t. $a$ and $b$).
3). Under the additional assumption that $n$ is a prime $p$, we have that $Br(E/F)= Br(E)[n]$ is a cyclic group of order $p$, and a non trivial class $(\chi, a)$ splits iff $a$ is a norm from $F(\sqrt [n] b)$. Suppose moreover that $F$ is a global field. Then CFT can be applied, and the so-called Hasse principle for cyclic extensions asserts that this global normic property holds iff all the local corresponding properties hold, i.e. the local Hilbert symbols $(a,b)_v$ are trivial for all the valuations $v$ of $F$ (actually this needs to be checked only for a finite number of $v$'s). If $p\neq 2$, there is no problem, one can take $a=b$ because the Hilbert symbol is anti-symmetric. If $p=2$, extra care is needed because the archimedean local Brauer groups are isomorphic to $\mathbf Z/2$. But they can be avoided since there is only a finite number of them. I skip this ./.