12$\cdot$102 Find all integer pairs $(x, y)$ that satisfy the equation $3 \cdot 2^{x}+1=y^{2}$.
(41st Moscow Mathematical Olympiad, 1978)
［Solution］When $x=0$,
$$
y^{2}=4, \quad y= \pm 2 .
$$

If $(x, y)$ is a solution to the given equation, then $(x,-y)$ is also a solution. Moreover, when $x<0$, the given equation has no integer solutions. Therefore, we only need to consider the positive integer solutions of the given equation.
Let $y=2 u+1$, where $u$ is a non-negative integer, then we have
$$
\begin{array}{l}
3 \cdot 2^{x}=4 u^{2}+4 u, \\
3 \cdot 2^{x-2}=u(u+1) .
\end{array}
$$

Since $(u, u+1)=1$, and one of $u$ and $u+1$ is odd while the other is even, it must be that
$$
\mathrm{I}\left\{\begin{array} { l } 
{ u = 3 , } \\
{ u + 1 = 2 ^ { x - 2 } , }
\end{array} \quad \text { II } \left\{\begin{array}{l}
u=2^{x-2}, \\
u+1=3 .
\end{array}\right.\right.
$$

From $\mathrm{I}$, we get $x=4, y=7$, and from II, we get $x=3, y=5$.
Thus, the given equation has solutions $(4,7),(4,-7),(3,5),(3,-5)$ and $(0,2),(0, -2)$, a total of six pairs.