Solve $2015+x^2=3^y4^z$

Question

If $x$, $y$ and z are positive integers, I want to solve $$2015+x^2=3^y4^z$$

What I tried:

I found that $x \equiv 1$ (mod $24$). So there exist an integer $p$ such that $24p+1=x$. Replacing that on the equation, we have $$48(12p^2+p+42)=3^y4^z$$