Example 4 Find all non-negative integer solutions $(x, y, z, w)$ of the indeterminate equation
$$
2^{x} \times 3^{y}-5^{x} \times 7^{w}=1
$$

[6]
From $5^{2} \times 7^{\infty}+1$ being even, we know $x \geqslant 1$.
(1) When $y=0$, $2^{x}-5^{x} \times 7^{w}=1$.

If $z \neq 0$, then $2^{x} \equiv 1(\bmod 5)$, thus, $4 \mid x$.
Therefore, $31\left(2^{x}-1\right)$, which contradicts $2^{x}-5^{x} \times 7^{x}=1$.
If $z=0,2^{x}-7^{w}=1$, when $x=1,2,3$, after inspection we get
$$
(x, w)=(1,0),(3,1) .
$$

When $x \geqslant 4$, we have $7^{w} \equiv-1(\bmod 16)$, direct calculation shows this is impossible.
Therefore, when $y=0$, all solutions are
$$
(x, y, z, w)=(1,0,0,0),(3,0,0,1) \text {. }
$$
(2) When $y>0, x=1$, $2 \times 3^{y}-5^{x} \times 7^{w}=1$.

Thus, $(-1)^{x} \equiv-1(\bmod 3)$. Therefore, $z$ is odd.
Hence, $2 \times 3^{y} \equiv 1(\bmod 5)$.
Therefore, $y \equiv 1(\bmod 4)$.
When $w \neq 0$, we have $2 \times 3^{y} \equiv 1(\bmod 7)$.
Therefore, $y \equiv 4(\bmod 6)$, which contradicts $y \equiv 1(\bmod 4)$.
So, $w=0$.
Thus, $2 \times 3^{y}-5^{z}=1$.
When $y=1$, $z=1$.
When $y \geqslant 2$, we have $5^{x} \equiv-1(\bmod 9)$.
Therefore, $z \equiv 0(\bmod 3)$.
Thus, $\left(5^{3}+1\right) \mid\left(5^{2}+1\right)$.
Hence $7 \mid\left(5^{x}+1\right)$, which contradicts $2 \times 3^{y}-5^{x}=1$.
The solution in this case is
$$
(x, y, z, w)=(1,1,1,0) \text {. }
$$
(3) When $y>0, x \geqslant 2$, we respectively take the original equation modulo 4 and modulo 3 to get
$$
(-1)^{w} \equiv-1(\bmod 4),(-1)^{z} \equiv-1(\bmod 3) .
$$

Therefore, $z$ and $w$ are both odd.
Thus, $2^{x} \times 3^{y}=5^{z} \times 7^{w}+1 \equiv 35+1 \equiv 4(\bmod 8)$.
Therefore, $x=2$.
The original equation becomes
$4 \times 3^{y}-5^{z} \times 7^{w}=1(z$ and $w$ are both odd $)$.
Thus, $4 \times 3^{y} \equiv 1(\bmod 5), 4 \times 3^{y} \equiv 1(\bmod 7)$.
From the above equations, we get $y \equiv 2(\bmod 12)$.
Let $y=12 m+2(m \geqslant 0)$. Then
$$
\begin{array}{l}
5^{z} \times 7^{w}=4 \times 3^{y}-1 \\
=\left(2 \times 3^{6 m+1}-1\right)\left(2 \times 3^{6 m+1}+1\right), \\
\text { and } 2 \times 3^{6 m+1}+1 \equiv 6 \times 2^{3 m}+1 \equiv 6+1 \equiv 0(\bmod 7), \\
\left(2 \times 3^{6 m+1}-1,2 \times 3^{6 m+1}+1\right)=1,
\end{array}
$$

and $\left(2 \times 3^{6 m+1}-1,2 \times 3^{6 m+1}+1\right)=1$,
then $51\left(2 \times 3^{6 m+1}-1\right)$.
Thus, $2 \times 3^{6 m+1}-1=5^{2}, 2 \times 3^{6 m+1}+1=7^{m}$.
If $m \geqslant 1$, then from the above equations we get $5^{x} \equiv-1(\bmod 9)$, which has already been proven impossible in case (2).
If $m=0$, then $y=2, z=1, w=1$.
The solution in this case is
$$
(x, y, z, w)=(2,2,1,1) \text {. }
$$

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