12.10. Solve the equation $2^{x}+7=y^{2}$ in natural numbers.
12.10. Answer: $x=1, y=3$. Let's check that there are no other solutions. It is clear that $2^{2}+7=11$ is not a square, so we can assume that $x \geqslant 3$. Then $2^{x}$ is divisible by 8, and therefore $y^{2} \equiv 7(\bmod 8)$. However, as it is easy to check, the square of an integer when divided by 8 can only give remainders 0, 1, and 4.