Is there an injective homomorphism between $\mathbb{Z_4} \times \mathbb{Z_4}$ and $S_7$?

Here’s a more conceptual solution which I think is cleaner. How many times is $7!$ divisible by $2$? A quick check yields $2^4$, so a $2$-Sylow subgroup of $S_7$ has order $16$.

Remember that all $2$-Sylow subgroups are conjugate, hence isomorphic. Your task, then, is to find such a group, and show that it is not isomorphic to $C_4\times C_4$, the product of two cyclic groups.

Constructing a group of order $16$ inside $S_7$ isn’t very hard: you should be familiar by now with dihedral groups. A dihedral group of size $8$ is easy to spot inside $S_7$, and then just multiply it by a cyclic group of order $2$ to obtain a Sylow subgroup. Finally, observe that this group isn’t isomorphic to $C_4\times C_4$ by counting elements of various orders (or, if you’re less silly than I am, by observing that one of these groups is abelian and the other is not.)