You're off to a good start. Here's how it goes.
$M$ is isomorphic to either $\mathbb Z_{21}$ or the Frobenius group, $F_{21}=\left<a,b\mid a^7=b^3=e,ba=a^2b\right>$. And clearly, $P_2\cong\mathbb Z_2$, say, $P_2=\left<c\right>$.
If $M\cong\mathbb Z_{21}$, say $M=\left<a\right>$, then $cac^{-1}=a^t$ for some $0\le t<21$ since $M$ is normal, thus $ca=a^tc$. Since $a=c^2a=ca^tc=a^{t^2}cc=a^{t^2}$, $t^2\equiv 1\pmod{21}$ so that $t$ is either $1$, $8$, $13$ or $20$. These four cases yield the groups $\mathbb Z_{42}$, $\mathbb Z_7\times S_3$, $D_7\times\mathbb Z_3$, and $D_{21}$ respectively.
Now suppose $M\cong F_{21}$. Then the map $\sigma:F_{21}\to F_{21}$ sending $x\to cxc^{-1}$ is an automorphism and $\sigma^2$ is the identity map $\iota$.
Now, $\sigma(a)=a^t$ for some $1\le t\le 6$ (because those are the only elements of $F_{21}$ of order $7$) and $\sigma^2=\iota$ implies $t^2\equiv 1\pmod 7$ so that $t$ is either $1$ or $6$. If $t=1$, then $\sigma(b)=a^ib$ for some $i$ (it can be shown that $\sigma(b)a=a^2\sigma(b)$ which would fail if $\sigma(b)=a^ib^2$); $\sigma^2=\iota$ implies $2i\equiv 0\pmod 7$, so that $\sigma(b)=b$ and $\sigma=\iota$. In this case, $M$ and $P_2$ commute, and $G\cong F_{21}\times\mathbb Z_2$.
Now suppose $t=6$. Then it doesn't matter what $\sigma(b)$ is; $\sigma^2=\iota$ would follow anyway, because if $\sigma(b)=a^ib$, then $\sigma(a^ib)=\sigma(a)^i\sigma(b)=(a^6)^ia^ib=a^{7i}b=b$. Hence, $\sigma^2(a)=a$ and $\sigma^2(b)=b$, so that $\sigma^2=\iota$. It can be shown that there are $7$ possibilities for $\sigma$ and each yields an isomorphic result for $G$, so suppose $\sigma(b)=b$. Then $bc=cb$, $\left<bc\right>$ is a cyclic subgroup of order $6$, $bca=a^5bc$ and $G$ is isomorphic to the holomorph $\mathbb Z_7\rtimes\mathbb Z_6$.
Summarizing, the six groups of order $42$ are $\mathbb Z_{42}$, $\mathbb Z_7\times S_3$, $D_7\times\mathbb Z_3$, $D_{21}$, $F_{21}\times\mathbb Z_2$ and the holomorph $\mathbb Z_7\rtimes\mathbb Z_6$. It can be shown quite feasibly that no two of those are isomorphic.
This is how I see it. $G$ has 3 2-Sylow subgroups, and 4 3-Sylow subgroups. So $G$ acts on the 3-Sylow subgroups by conjugation, resulting in a map $\phi:G → S_4$. Let $K$ be the kernel of this map. Let $P$ be a 3-Sylow subgroup, and restrict $\phi$ to P; call the map $\phi_P$. Then the Sylow Permutation Theorem* states that the only fixed point of $\phi_P$ on the 4 conjugates is $P$ itself. This means that $\phi_P$ acts nontrivially on the other three 3-Sylow subgroups, implying that $K$ cannot contain an element of order 3, so that the order of $K$ is not divisible by 3, so it has to be 1, 2, 4, or 8.
Let $\psi: G → S_3$ be the action of $G$ on the 2-Sylow subgroups. Then $G$, order 24, acts on $S_3$ of order 6, implying a normal subgroup $V$ in $G$ of order 4. This subgroup then combines with a 3-Sylow subgroup, by semi-direct product, to form a subgroup $A$ of $G$ of order 12. There can be only one such subgroup of order 12, as there is no room for any more. So $A$ is of order 12 without normal 3-Sylow subgroups, so it is isomorphic to $A_4$ and $V$ is isomorphic to the Klein four-group. The 2-Sylow subgroups are of order 8, and each one of them must contain $V$ as a subgroup. Hence we have 1 identity element, 3 elements in $V$, 4 x 3 or 12 elements in the 2-Sylows outside of $V$, and 8 elements of order 3, accounting for all 24 elements of G. There is no more room for any more.
$K$ cannot be of order 8, because the subgroups of order 8 are not normal. $K$ cannot be of order 2, since in that case $K$ would be normal in $G$, so it combines with $P$ to form a subgroup of order 6 with subgroups of order 2 which are normal, hence is $Z_6$, containing elements of order 6, for which there is no room. Finally, $K$ can't be of order 4, since the normalizer of $P$ and its conjugates is of order 6, so the intersection of all these normalizers has to have order 1, 2, 3, or 6 and be contained in $K$, so it can't have order 4.
Therefore, $K$ is the trivial group, so that $\phi$ is an isomorphism between $G$ and $S_4$.
*The Sylow Permutation Theorem says that if a group $G$ has $n$ $p$-Sylow subgroups, then $G$ acts on $Syl_p(G)$ by conjugation, and the only fixed point of this action restricted to $P$ is $P$ itself. This occurs throughout the literature, mainly in proving that there are no simple groups of some specified order, but not as a theorem by itself.
Best Answer
The proof is good. I think it can be structured more clearly and the groups can be explicitly identified. We know that the following groups exist:
Sylow theory tells us that the Sylow 3-subgroups will be $C_3$, and the Sylow 2-subgroups will be $C_4$ or $C_2 \times C_2$. We also learn that:
When $n_2 = n_3 = 1$ we have the Abelian groups.
When $n_3 = 4$ you showed that we have $A_4$.
We can now look at the only remaining case: $n_3 = 1$ and $n_2 = 4$. In this situation we are searching for the nontrivial semidirect products $C_3 \rtimes_\theta P_2$ with $\theta : P_2 \to \operatorname{Aut}(C_3)$.
Note that $\operatorname{Aut}(C_3) \simeq \langle id, inv \rangle \simeq C_2$
Let's split into cases based on what $P_2$ is.
(Case A) $P_2 = C_4$:
In this case there is exactly one nontrivial homomorphism which is forced from $\theta(0) = 0$ and $\theta(1) = 1$. This gives us the metacyclic group, $Dic_3$.
(Case B) $P_2 = C_2 \times C_2$:
In this case there are 3 different nontrivial homomorphisms:
Now these's actually all give us isomorphic semidirect products because we have automorphisms of $P_2$ that relate these maps to each other:
Now we can use $\theta_a$ and the definition of multiplication of elements in a semidirect product to see that $C_3 \rtimes_{\theta_a} C_2 \times C_2 \simeq S_3 \times C_2 \simeq D_{6}$.