6 Find all non-negative integer solutions $(x, y, z, w)$ of the equation
$$
2^{x} \cdot 3^{y}-5^{x} \cdot 7^{w}=1
$$
Given that $5^{x} \cdot 7^{w}+1$ is even, we know $x \geqslant 1$.
i If $y=0$, then
$$
2^{x}-5^{x} \cdot 7^{w}=1
$$

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

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

This 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 $y=1(\bmod 4)$. Therefore, $w=0$, and we have
$$
2 \cdot 3^{y}-5^{z}=1
$$

When $y=1$, $z=1$. When $y \geqslant 2$, we have $5^{z} \equiv-1(\bmod 9)$, which implies $z \equiv 3(\bmod 6)$. Therefore, $5^{3}+115^{2}+1$, hence $715^{2}+1$, which contradicts $5^{z}+1=2 \cdot 3^{y}$. Thus, the solution in this case is
$$
(x, y, z, w)=(1,1,1,0)
$$
iii If $y>0, x \geqslant 2$, then
$$
5^{z} \cdot 7^{w} \equiv-1(\bmod 4), 5^{z} \cdot 7^{w} \equiv-1(\bmod 3)
$$

Thus,
$$
(-1)^{w} \equiv-1(\bmod 4),(-1)^{z}=-1(\bmod 3)
$$

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

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

This implies $4 \cdot 3^{y} \equiv 1(\bmod 5), 4 \cdot 3^{y} \equiv 1(\bmod 7)$
From the above two equations, we get $y=2(\bmod 12)$.
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 since $\left(2 \cdot 3^{6 m+1}-1,2 \cdot 3^{6 m+1}+1\right)=1$
Therefore, $512 \cdot 3^{6 m+1}-1$, thus
$$
\begin{array}{l}
2 \cdot 3^{6 m+1}-1=5^{2} \\
2 \cdot 3^{6 m+1}+1=7^{w}
\end{array}
$$

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

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