Nice diophantine! Answer: No solution We have $$5^x7^y+4=3^z.$$ By $\mod 3,$ we get that $x$ is odd. By $\mod 4,$ we get that $y,z$ are of the same parity. By $\mod 7,$ we get that $z$ is even and so is $y.$ Let $z=2k.$ Hence, $$5^x\cdot 7^y= (3^k-2)(3^k+2).$$Since $$\gcd(3^k-2),(3^k+2)=1\implies 3^k-2=5^x \text{or} 3^k+2=5^x$$ Case 1: $$ 3^k-2=5^x$$Clearly $k>1,$ taking $\mod 9$ and using the fact that $x$ is odd, so $5^x\equiv 5,8,2 \mod 9.$ Not possible ( except $k=1,x=0$ which eventually doesn't work when we try it in $3^k+2=7^y.$ ) (another way is just $\mod 3$) Case 2: $$3^k+2=5^x, 3^k-2=7^y$$Clearly in $3^k+2=5^x,$ we get by $\mod 8 \implies k$ is odd. But $\mod 7$ in $3^k-2=7^y$ gives $k$ is even. Not possible.