6. The system of equations
$$
\left\{\begin{array}{l}
(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)=1+y^{\prime} \\
(1+y)\left(1+y^{2}\right)\left(1+y^{1}\right)=1+x^{7}
\end{array}\right.
$$

has ( ) solutions for $(x, y)$.
(A) 4
(B) 2
(C) 1
(D) 0
6. (B).

Given the equations can be transformed into
$$
\left\{\begin{array}{l}
1-x^{8}=(1-x)\left(1+y^{7}\right), \\
1-y^{8}=(1-y)\left(1+x^{7}\right) .
\end{array}\right.
$$

When $x=y$, it is clear that $x=1$. This can be solved to get
$$
(x, y)=(C, 0),(-1, \cdots 1) \text {. }
$$

When $x \neq y$,
if $x>0, y>0$, then
$$
(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)>1+x^{7}>1+y^{7},
$$

the equation does not hold.
If $0>x>y$, then $x^{7}>y^{7}, x^{6}<y^{6}, x^{8}<y^{8}$.
Subtracting (2) from (1) gives
$$
\begin{aligned}
0 & <y^{8}-x^{8}=(y-x)+\left(y^{7}-x^{7}\right)+x y\left(x^{6}-y^{6}\right) \\
& <0 .
\end{aligned}
$$

This is a contradiction.
Therefore, the system of equations has no solutions for $x \neq y$, meaning there are only two solutions.