Let $E$ be the splitting field of $x^{42}-1$ over $\mathbb{Q}$.
Idea 1: $x^{42}-1=0\iff x=e^{\pi ik/21}, k=0,\ldots, 41.$ So $E=\mathbb{Q}(e^{\pi i/21})$. I know that $\mathbb{Q}(e^{\pi ik/7}), \mathbb{Q}(e^{\pi i/3})$ are nontrivial subfields of $E$, but how can I determine whether they are the only ones?
Idea 2: Now $$x^{42}-1=\Phi_1(x)\Phi_2(x)\Phi_3(x)\Phi_6(x)\Phi_7(x)\Phi_{14}(x)\Phi_{21}(x)\Phi_{42}(x)$$ But I am not sure how to use this since I don't know if, say, containing primitive $3$rd, $7$th roots of unity implies we have $21$st roots of unity over a field of characteristic $0$.
The splitting field of $x^{42} - 1$ over $\mathbb{Q}$ is a Galois extension, with Galois group isomorphic to the multiplicative group of integers modulo $42$
This is so because with $\zeta$ a primitive $42^{\text{nd}}$ root of unity, any $\sigma \in \text{Gal}(E/\mathbb{Q})$ must send $\zeta$ to another primitive root of unity, i.e. $\zeta^k$ for some $k$ coprime to $42$.
Now mapping $\sigma \mapsto k$ is a homomorphism from $\text{Gal}(E/\mathbb{Q})$ to the multiplicative group modulo $42$. Furthermore, since the $42^{\text{nd}}$ cyclotomic polynomial is irreducible over $\mathbb{Q}$, this homomorphism is bijective.
All of the above is standard theory - see e.g. Cyclotomic Extensions
Application to your question
The Fundamental Theorem of Galois Theory implies that there is a 1-1 correspondence between subgroups of $\text{Gal}(E/\mathbb{Q})$ and the intermediate fields $\mathbb{Q} \subset F \subset E$.
Can you find all the subgroups of $\left( \mathbb{Z} / 42\mathbb{Z}\right)^{*}$? This will provide you with your answer.