2 $*$ Find all non-negative integer solutions $(x, y, z, w)$ of the equation
$$
2^{x} \cdot 3^{y}-5^{z} \cdot 7^{w}=1
$$
Parse $5^{z} \cdot 7^{w}+1$ as an even number, so $x \geqslant 1$.
(1) $y=0$. (1) becomes $2^{x}-1=5^{z} \cdot 7^{w}$.
(2) If $z \neq 0$, then $5 \mid 2^{x}-1, x$ is a multiple of 4. But at this time $2^{x}-1$ is divisible by 3, which contradicts (2). So $z=0$. (2) becomes $2^{x}-1=7^{w}$. When $x=1$, $w=0$. When $x=2$, there is no solution. When $x=3$, $w=1$. When $x \geqslant 4$, $7^{w} \equiv 1(\bmod 16)$. But $7^{1} \equiv 7,7^{2} \equiv 1(\bmod 16)$, so $7^{w} \not \equiv-1(\bmod 16)$.
(2) $y \geqslant 1, x=1$. At this time $-(-1)^{z} \equiv 1(\bmod 3)$, so $z$ is odd. $2 \cdot 3^{y} \equiv 1(\bmod 5)$, so $y \equiv 1(\bmod 4)$. If $w \neq 0$, then $2 \cdot 3^{y} \equiv 1(\bmod 7)$, so $y \equiv 4(\bmod 6)$, $y$ is even. This contradicts $y=1(\bmod 4)$. Therefore $w=0$. (1) becomes $2 \cdot 3^{y}-5^{z}=1$. When $y=1$, $z=1$. If $y \geqslant 2$, then $5^{z} \equiv-1(\bmod 9)$. Thus $z \equiv 3(\bmod 6)$. At this time $5^{z}+1$ is divisible by $5^{3}+1-7 \times 9 \times 2$, which contradicts $5^{z}+1-2 \cdot 3^{y}$ not being divisible by 7.
(3) $y \geqslant 1, x \geqslant 2$. As in (2), $z$ is odd. Also $-7^{w} \equiv 1(\bmod 4)$, so $w$ is odd. $2^{x} \cdot 3^{y}=5^{z} \cdot 7^{w}+1=5 \cdot 7+1=4(\bmod 8)$, so $x=2, y$ is even $2k$.
$5^{z} \cdot 7^{w}=4 \cdot 3^{2k}-1=\left(2 \cdot 3^{k}+1\right)\left(2 \cdot 3^{k}-1\right)$.
$2 \cdot 3^{k}+1$ and $2 \cdot 3^{k}-1$ are both odd, coprime, and the former $\equiv 1(\bmod 3)$, so $2 \cdot 3^{k}+1=7^{w}$, $2 \cdot 3^{k}-1=5^{z}$.
If $k \geqslant 2$, as above, $5^{z}+1$ is divisible by 7, which is impossible. So $k=1, z=1, w=1$. The solutions to this problem are 4 sets, namely $(x, y, z, w)=(1,0,0,0),(3,0,0,1),(1,1,1,0),(2,2,1,1)$.