8. Find the positive integer solutions of the equation $3 \cdot 2^{x}+1=y^{2}$.

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8. When $x=1$ or 2, $y$ has no integer solutions. Therefore, $x>2$, $y$ is odd and not divisible by 3. Hence, we can set $y=6k \pm 1, k \in N$, then $2^{x-2}=k(3k \pm 1)$. It is easy to see that $k=1$, thus we have $y=5, x=3$ or $y$ $=7, x=4$.