Proving that a group of order $112$ is not simple
If $G$ is a simple group, it must have exactly $7$ Sylow $2$-subgroups. Thus $G$ embeds into $S_7$, and in particular into $A_7$ since $G$ does not have a subgroup of index $2$. But the order $A_7$ is not divisible by $112$.
If you want to go along the lines of your original idea, you can rule out the case $|P \cap Q| = 2^3$ by noticing that then $P \cap Q$ is normal in $P$ and $Q$ (as a subgroup of index $2$), so $N_G(P \cap Q)$ contains $P$ and $Q$, which implies that $N_G(P \cap Q) = G$.
ADDED: I'm not sure if there is an easy way to deal with rest of the cases. However, there is a nice argument which also works for proving that every group of order $p^n q$ ($p$, $q$ distinct primes) is nonsimple. I believe the idea of the proof goes back to G. A. Miller (around 1900-1910). Here's an illustration of it in this case.
Suppose that $G$ is a simple group of order $112$. Then $G$ has exactly $7$ Sylow $2$-subgroups. Let $P, Q \in Syl_2(G)$ be such that $P \neq Q$ and that $D = P \cap Q$ has largest possible order. Steps for the proof:
Using the fact that $D < N_P(D)$ and $D < N_Q(D)$ (proper inclusion), prove that $N_G(D)$ cannot be a $2$-group.
Thus $D$ is normalized by an element $g \in G$ of order $7$. Prove that $P, gPg^{-1}, \ldots, g^6Pg^{-6}$ are distinct. Conclude that $D$ is contained in every Sylow $2$-subgroup.
Since the intersection of all Sylow $2$-subgroups is normal, $D$ is trivial.
By counting elements in Sylow $2$-subgroups, prove that $G$ contains exactly one Sylow $7$-subgroup.
This same argument works for proving the statement for groups of order $p^n q$.
I wanted to write out Mikko Korhonen's first idea for proof in detail as a separate answer, since it is not trivial at all, and provoked some questions in the comments.
As mentioned in the original question, we assumed $n_2=7$. From Sylow's second theorem, we know that all the 2-Sylow subgroups are conjugate, and we can look at $G$'s action on them by conjugation. This action implies a homomorphism: $$f:G\rightarrow S_7$$ $\ker(f)$ cannot be $G$ since the action is a transitive action. Then, if $\ker(f)$ is non-trivial (meaning $ker(f)\neq\{e\}$) then it is a non-trivial normal subgroup of $G$, and therefore $G$ isn't simple.
Otherwise, we get $ker(f)=\{e\}$ and therefore $f$ is injective, meaning that $G$ is isomorphic to a subgroup of $S_7$. For convenience purposes, we'll write $G\leq S_7$. $G$ cannot be contained in $A_7$ because 112 doesn't divide $|A_7|$. In that case $GA_7=S_7$ and using the second isomorphism theorem we get: $$G/(G\cap A_7)\cong GA_7/A_7\cong S_7/A_7\cong\mathbb{Z}_2$$ and therefore $[G:(G\cap A_7)]=2$ meaning that $G\cap A_7$ is a normal subgroup of $G$.