6. If $x, y$ are positive integers, then the number of positive integer solutions $(x, y)$ for the equation $15 \times 2^{x}+1=$ $y^{2}$ is $(\quad)$.
(A) 0
(B) 1
(C) 2
(D) 4
6. C.

Obviously, $y$ is an odd number greater than 1.
Let $y=2k+1\left(k \in \mathbf{N}_{+}\right)$. Then $15 \times 2^{x-2}=k(k+1)$.
Note that $k$ and $k+1$ are one odd and one even.
(1) When $k$ is odd, $k=3,5,15$. By calculation, when $k=3$, there is no solution; when $k=5$, $x=3, y=11$; when $k=15$, $x=6, y=31$.
(2) When $k$ is even, $k+1=3,5,15$, i.e., $k=2,4,14$. By calculation, none of them meet the requirements of the problem.
In summary, the positive integer solutions $(x, y)$ of the equation are 2 in total.