Find all integers $x, y, z \geq 0$ such that $5^{x} 7^{y}+4=3^{z}$.
It is clear that $(x, y) \neq(0,0)$ and that $z \geq 1$. First, $(x, y, z)=(1,0,2)$ is indeed a solution.

If $y=0$ (and $x \geq 1$), by looking at the equation modulo 5, we find $z \equiv 2 \bmod 4$. If $y \geq 1$, by looking modulo 7, we find that $z \equiv 4 \bmod 6$. In all cases, $z$ is even. We can therefore write $z=2 n$.

Then $5^{x} 7^{y}=\left(3^{n}-2\right)\left(3^{n}+2\right)$. Since $3^{n}-2$ and $3^{n}+2$ are coprime, two cases arise:

Case 1: $3^{n}-2=5^{x}$ and $3^{n}+2=7^{y}$. By looking at the second equation modulo 3, we see that there are no solutions.

Case 2: $3^{n}-2=7^{y}$ and $3^{n}+2=5^{x}$. Then $4=5^{x}-7^{y}$. Suppose, for the sake of contradiction, that $x, y \geq 1$. By looking modulo 7, we see $x \equiv 2 \bmod 6$ and thus $x$ is even. By looking modulo 5, we see that $y \equiv 0 \bmod 4$ and thus $y$ is even. By writing $x=2 x^{\prime}$ and $y=2 y^{\prime}$, we have $4=\left(5^{x^{\prime}}-7^{y^{\prime}}\right)\left(5^{x^{\prime}}+7^{y^{\prime}}\right)$, which does not yield solutions. Therefore, we have $x=0$ or $y=0$, and we find the only solution $(x, y, z)=(1,0,2)$.