6 Find all non-negative integer solutions $(x, y, z, w)$ to the equation
$$
2^{x} \cdot 3^{y}-5^{z} \cdot 7^{w}=1
$$

(Supplied by Yonggao Chen)
Given that $5^{z} \cdot 7^{w}+1$ is even, we know $x \geqslant 1$.

Case 1: If $y=0$, then
$$
2^{x}-5^{z} \cdot 7^{w}=1
$$

If $z \neq 0$, then $2^{x} \equiv 1(\bmod 5)$, which implies $4 \mid x$. Therefore, $3 \mid 2^{x}-1$, which contradicts $2^{x}-5^{z} \cdot 7^{w}=1$.
If $z=0$, then
$$
2^{x}-7^{w}=1
$$

When $x=1,2,3$, direct calculation yields $(x, w)=(1,0),(3,1)$. When $x \geqslant 4$, we have $7 w \equiv-1(\bmod 16)$, which is impossible by direct calculation. Therefore, when $y=0$, all non-negative integer solutions are
$$
(x, y, z, w)=(1,0,0,0),(3,0,0,1).
$$

Case 2: If $y>0, x=1$, then
$$
2 \cdot 3^{y}-5^{z} \cdot 7^{w}=1
$$

Thus, $-5^{z} \cdot 7 w \equiv 1(\bmod 3)$, i.e., $(-1)^{z} \equiv-1(\bmod 3)$, so $z$ is odd, hence
$$
2 \cdot 3^{y} \equiv 1(\bmod 5),
$$

which implies $y \equiv 1(\bmod 4)$.
When $w \neq 0$, we have $2 \cdot 3^{y} \equiv 1(\bmod 7)$, thus $y \equiv 4(\bmod 6)$, which contradicts $y \equiv 1(\bmod 4)$. Therefore, $w=0$. Thus,
$$
2 \cdot 3^{y}-5^{z}=1
$$

When $y=1$, $z=1$. When $y \geqslant 2$, we have $5^{z} \equiv-1(\bmod 9)$, which implies $z \equiv 3(\bmod 6)$, so $5^{3}+1 \mid 5^{z}+1$, hence $7 \mid 5^{z}+1$, which contradicts $5^{z}+1=2 \cdot 3^{y}$. Therefore, the solution in this case is
$$
(x, y, z, w)=(1,1,1,0).
$$

Case 3: If $y>0, x \geqslant 2$, then
$$
5^{z} \cdot 7^{w} \equiv-1(\bmod 4), \quad 5^{z} \cdot 7^{w} \equiv-1(\bmod 3),
$$

i.e.,
$$
(-1)^{w} \equiv-1(\bmod 4), \quad(-1)^{z} \equiv-1(\bmod 3),
$$

so $z$ and $w$ are both odd, thus
$$
2^{x} \cdot 3^{y}=5^{z} \cdot 7^{w}+1 \equiv 35+1 \equiv 4(\bmod 8),
$$

so $x=2$, the original equation becomes
$$
4 \cdot 3^{y}-5^{z} \cdot 7^{w}=1 \text{ (where } z \text{ and } w \text{ are both odd). }
$$

Thus,
$$
4 \cdot 3^{y} \equiv 1(\bmod 5), \quad 4 \cdot 3^{y} \equiv 1(\bmod 7),
$$

from which we get $y \equiv 2(\bmod 12)$.
Let $y=12 m+2, m \geqslant 0$, then
$$
5^{z} \cdot 7^{w}=4 \cdot 3^{y}-1=\left(2 \cdot 3^{6 m+1}-1\right)\left(2 \cdot 3^{6 m+1}+1\right).
$$

Since
$$
2 \cdot 3^{6 m+1}+1 \equiv 6 \cdot 2^{3 m}+1 \equiv 6+1 \equiv 0(\bmod 7),
$$

and $\left(2 \cdot 3^{6 m+1}-1,2 \cdot 3^{6 m+1}+1\right)=1$, so $5 \mid 2 \cdot 3^{6 m+1}-1$, thus
$$
\begin{array}{l}
2 \cdot 3^{6 m+1}-1=5^{z} \\
2 \cdot 3^{6 m+1}+1=7 w
\end{array}
$$

If $m \geqslant 1$, then from (2) we get $5^{z} \equiv-1(\bmod 9)$, which is impossible by Case 2.
If $m=0$, then $y=2, z=1, w=1$.
Therefore, the solution in this case is
$$
(x, y, z, w)=(2,2,1,1).
$$

In summary, the non-negative integer solutions are
$$
\begin{aligned}
(x, y, z, w)= & (1,0,0,0),(3,0,0,1), \\
& (1,1,1,0),(2,2,1,1).
\end{aligned}
$$