Suppose that $G$ is a group of order $924=2^2\cdot3\cdot7\cdot 11$. Prove that $G$ has an element of order $77$.

Well we know there are elements of order $7$ and $11$ and if any pair of such elements commute then they generate a cyclic subgroup of order $77$.

I think you can argue that if the number of subgroups of order $11$ is not $1$ then it is $12$ (Sylow again: $\equiv 1 \bmod 11$). Take these two cases together.

Take an element $a$ of order $7$ and let it act on these subgroups by conjugation. The orbits must either be single subgroups or sets of $7$ subgroups. In either case there is a subgroup of order $11$ fixed under the conjugation action.

Now consider the action on that subgroup - the automorphism group of a cyclic group of order $11$ has order $10$ and the action induces a homomorphism from the group of order $7$ generated by $a$ to the automorphism group. The image is a subgroup of order $1$ or $7$, and it must be $1$. Therefore $a$ acts trivially on the subgroup of order $11$ and commutes with its members.

You can bypass Burnside's lemma completely.

The number of Sylow-11's is $n_{11}=1$ or $n_{11}=12$. Since all Sylow-11s are conjugate, we get $n_{11}=\lvert G\rvert/\lvert N(P_{11})\rvert$ from orbit-stabilizer (as in proof of Sylow first theorem), i.e., $\lvert N(P_{11})\rvert=\lvert G\rvert/ n_{11}$, which is divisible by $7$ in both cases. So there is a Sylow-7 normalizing a Sylow-11, i.e., there is a $C_{11}\rtimes C_7$ as a subgroup of $G$. But there no nontrivial homomorphism $C_7\to\operatorname{Aut}C_{11}$ so this is just $C_{77}$.