6. The correct judgment about the positive integer solutions of the equation $1+2+\cdots+x=y^{2}$ is ( ).
(A) None
(B) 1
(C) More than 1 but finite
(D) Infinite
6.D.

It is easy to know that $1+2+\cdots+8=\frac{8 \times 9}{2}=6^{2}$. Therefore, $x=8, y=6$ is a positive integer solution to the equation.
Assume $\left(x_{k}, y_{k}\right)$ is a positive integer solution to the equation, i.e.,
$$
1+2+\cdots+x_{k}=\frac{x_{k}\left(x_{k}+1\right)}{2}=y_{k}^{2} \text {, }
$$

then $1+2+\cdots+\left[\left(2 x_{k}+1\right)^{2}-1\right]$
$=\frac{\left[\left(2 x_{k}+1\right)^{2}-1\right]\left(2 x_{k}+1\right)^{2}}{2}$
$=4 \times \frac{x_{k}\left(x_{k}+1\right)}{2}\left(2 x_{k}+1\right)^{2}$
$=4 y_{k}^{2}\left(2 x_{k}+1\right)^{2}=\left(4 x_{k} y_{k}+2 y_{k}\right)^{2}$.
This shows that $x_{k+1}=\left(2 x_{k}+1\right)^{2}-1, y_{k+1}=4 x_{k} y_{k}+$ $2 y_{k}$, is also a positive integer solution to the equation.
Repeating this process, we can see that the equation has infinitely many positive integer solutions.