China Mathematical Olympiad

Since $5^z \cdot 7^w + 1$ is even, we have $x \ge 1$.

Case 1: $y = 0$. The equation to be solved becomes
$$2^x - 5^z \cdot 7^w = 1.$$
If $z \neq 0$, then $2^x \equiv 1 \pmod{5}$. It follows that $4 \mid x$. Thus $3 \mid 2^x - 1$, which contradicts to $2^x - 5^z \cdot 7^w = 1$.
If $z = 0$, then
$$2^x - 7^w = 1.$$
When $x = 1, 2, 3$, a direct computation shows that $(x, w) = (1, 0), (3, 1)$ are the solutions.
When $x \ge 4$, $7^w \equiv -1 \pmod{16}$. By direct computation we know that this is impossible.
Consequently, when $y = 0$ all non-negative integer solutions of the equation are
$$(x, y, z, w) = (1, 0, 0, 0), (3, 0, 0, 1).$$

Case 2: $y > 0$ and $x = 1$. Thus the equation to be solved becomes
$$2 \cdot 3^y - 5^z \cdot 7^w = 1.$$
Hence $-5^z \cdot 7^w \equiv 1 \pmod{3}$, i.e., $(-1)^z \equiv -1 \pmod{3}$. It follows that $z$ is odd.
Thus
$$2 \cdot 3^y \equiv 1 \pmod{5}.$$
$$y \equiv 1 \pmod{4}.$$
When $w \neq 0$, we have $2 \cdot 3^y \equiv 1 \pmod{7}$. Thus $y \equiv 4 \pmod{6}$, which contradicts to the fact $y \equiv 1 \pmod{4}$. Hence $w = 0$ and
$$2 \cdot 3^y - 5^z = 1.$$
When $y = 1$, we have $z = 1$. If $y \ge 2$, then $5^z \equiv -1 \pmod{9}$, which implies $z \equiv 3 \pmod{6}$. Thus $5^3 + 1 \pmod{5^z + 1}$, so $7 \pmod{5^z + 1}$, which contradicts to $5^z + 1 = 2 \cdot 3^y$. Hence in this case we have only one solution
$$(x, y, z, w) = (1, 1, 1, 0).$$

Case 3: $y > 0$ and $x \ge 2$. Thus
$$5^z \cdot 7^w \equiv -1 \pmod{4}, \text{ and } 5^z \cdot 7^w \equiv -1 \pmod{3}.$$
That is,
$$(-1)^w \equiv -1 \pmod{4}, \text{ and } (-1)^z \equiv -1 \pmod{3}.$$
Thus $z$ and $w$ are odd. It follows that
$$2^x \cdot 3^y = 5^z \cdot 7^w + 1 \equiv 35 + 1 \equiv 4 \pmod{8}.$$
Hence, $x = 2$, and
$$4 \cdot 3^y - 5^z \cdot 7^w = 1 \text{ (where $z$ and $w$ are odd).}$$
Thus,
$$4 \cdot 3^y \equiv 1 \pmod{5}, \text{ and } 4 \cdot 3^y \equiv 1 \pmod{7}.$$
From the above two congruencies we have $y \equiv 2 \pmod{12}$.
Set $y = 12m + 2$, $m \ge 0$, then
$$5^z \cdot 7^w = 4 \cdot 3^y - 1 = (2 \cdot 3^{6m+1} - 1)(2 \cdot 3^{6m+1} + 1).$$
Since
$$2 \cdot 3^{6m+1} + 1 \equiv 6 \cdot 2^{3m} + 1 \equiv 6 + 1 \equiv 0 \pmod{7},$$
and
$$(2 \cdot 3^{6m+1} - 1, 2 \cdot 3^{6m+1} + 1) = 1,$$ so $5 \mid 2 \cdot 3^{6m+1} - 1$.
Thus
$$2 \cdot 3^{6m+1} - 1 = 5^z,$$
$$2 \cdot 3^{6m+1} + 1 = 7^w.$$
If $m \ge 1$, by Equation (2) we have $5^z \equiv -1 \pmod{9}$, and from Case 2 we know that this is impossible.
If $m=0$, then $y=2$, $z=1$ and $w=1$. Thus in this case, we have only one solution
$$(x, y, z, w) = (2, 2, 1, 1).$$

Consequently, all non-negative integer solutions are
$$(x, y, z, w) = (1, 0, 0, 0), (3, 0, 0, 1), \\ (1, 1, 1, 0), (2, 2, 1, 1).$$