First of all notice that $x, x+1, x+2, x+3$ all have a different reminder $\pmod{4}$ hence one of $ (x+1)(x+2)(x+3), x(x+2)(x+3), x(x+1)(x+3), x(x+1)(x+2)$ must not divide 4 and the other ones do, thus $ (x+1)(x+2)(x+3) +x(x+2)(x+3)+x(x+1)(x+3)+x(x+1)(x+2)\equiv 1\cdot 2\cdot 3\equiv 2 \pmod{4}$ but $y^{2^x}\equiv 1 or 0 \pmod{4}$ unless $x=0$. So $x=0$, this means that $y=6$. If $x<0$, then $y^{2^x}\not\in \mathbb{Z}$ whilst $LHS\in \mathbb{Z}$ which is a contradiction. Hence the only solution is $x=0$ and $y=6$.