The right side is $2^{x+y}\left(1+2^{x+y-1}\right)$. If the second factor is odd, it needs to be a power of 3 , so the only options are $x+y=2$ and $x+y=4$. This leads to two solutions, namely $(1,1)$ and $(2,2)$. The second factor can also be even, if $x+y-1=0$. Then $x+y=1$ and $3^{x} 4^{y}=2+2$, giving $(0,1)$ as the only other solution.