3. The set of positive integer solutions $(x, y)$ for the indeterminate equation $3 \times 2^{x}+1=y^{2}$ is $\qquad$
3. $\{(3,5),(4,7)\}$.

Obviously, $y$ is odd. Taking both sides of the equation modulo 8, we know $x \geqslant 3$. Transforming the original equation, we get
$$
3 \times 2^{x-2}=\frac{y-1}{2} \cdot \frac{y+1}{2} \text {. }
$$

Since $\left(\frac{y-1}{2}, \frac{y+1}{2}\right)=1$, we have $x=3$ or 4. Therefore, the original equation has exactly two solutions $(3,5),(4,7)$.