There is a following generalization of one of the Sylow's theorems (see for example here), which is equivalent to the theorem in Cihan's answer.
Let $P_1, P_2, \ldots, P_k$ be the Sylow $p$-subgroups of $G$. If $[P_i : P_i \cap P_j] \geq p^d$ whenever $i \neq j$, then $n_p \equiv 1 \mod p^d$.
Taking $d = 1$ gives you one of Sylow's theorems, and the contrapositive of this theorem in the case $d = 2$ is useful here. If $n_p \not\equiv 1 \mod p^2$, then there exist different Sylow $p$-subgroups $P$ and $Q$ with $[P: P \cap Q] < p^2$. This implies that $[P : P\cap Q] = [Q : P \cap Q] = p$, and thus the intersection $P \cap Q$ is normal in both $P$ and $Q$.
In your problem, we can assume $n_2 = 3$. Then $3 \not\equiv 1 \mod 4$, so we find different Sylow $2$-subgroups $P$ and $Q$ with $P \cap Q$ normal in both $P$ and $Q$. Therefore the intersection $P \cap Q$ is normal in the subgroup $\langle P, Q \rangle$ generated by $P$ and $Q$. The order of $ \langle P, Q \rangle$ is a multiple of $2^5$ and larger than $2^5$, so it has to equal $G$. Thus $P \cap Q$ is a nontrivial proper normal subgroup of $G$.