12. Answer: 6
Note that $x$ and $y$ must satisfy
$$
2^{\pi+1} \cdot 3=y(y+2)
$$

We first assume $x \geq 0$, which means both $y$ and $y+2$ are even integers. Either $3 y$ or $3 y+2$. In the first case, assuming $y>0$, we have $y=3 \cdot 2^{k}$ and $y+2=2\left(3 \cdot 2^{k-1}+1\right)=2^{x+1-k}$. The only way for this equation to hold is $k=1$ and $x=3$. So $(x, y, y+2)=(3,6,8)$.
In the case $3 y+2$, assuming $y>0$, we have $y+2=3 \cdot 2^{k}$ and $y=2\left(3 \cdot 2^{k-1}-1\right)=$ $2^{x+1-k}$. Now the only possibility is $k=1$ and $x=2$, so $(x, y, y+2)=(2,4,6)$.
In the two previous cases, we could also have both $y$ and $y+2$ to be negative, giving $(x, y, y+2)=(3,-8,-6)$ or $(2,-6,-4)$.
Finally, we consider $x<0$ so $3=2^{-x-1} y(y+2)$. In this case we can only have $x=-1$ and $(x, y, y+2)=(-1,1,3)$ or $(-1,-3,-1)$.
Hence possible $(x, y)$ pairs are $(3,6),(3,-8),(2,4),(2,-6),(-1,1)$ and $(-1,-3)$.
12. Determine the total number of pairs of integers $x$ and $y$ that satisfy the equation
$$
\frac{1}{y}-\frac{1}{y+2}=\frac{1}{3 \cdot 2^{x}}
$$