Solve in positive integers: $5^x 7^y +4=3^z$ Solve in positive integers: $5^x 7^y +4=3^z$.
I tried to solve it with log but I couldn't complete.

[Based on the work from @Adam Hughes]
We assume $x>1$ and rewrite it:
$$3^z-4=5^x7^y \Rightarrow
3^z-4\equiv 0\mod 5 \iff 3^z\equiv -1\mod 5$$
so $z=2j$ is even. 
Then we have $(3^j-2)(3^j+2)=5^x7^y$. Since the gcd of these two divides $4$--because $(3^j+2)-(3^j-2)=4$--we have that they are coprime (any larger factor would then be a factor of $5^x7^y$ which is odd), i.e. one of the following two cases hold
$$\begin{cases} 3^j+2=5^x\quad ,\quad 3^j-2=7^y \\ 3^j+2=7^y\quad ,\quad3^j-2=5^x\end{cases}.$$
Case 2:
$3^j+2=7^y\quad ,\quad3^j-2=5^x \Rightarrow 7^y = 4 + 5^x$, which is impossible if you look at it $\mod 3$.
So we have Case 1:
$3^j-2=7^y\quad ,\quad3^j+2=5^x \Rightarrow 7^y+4 = 5^x$. If you look at it $\mod 4$, you get $2|y$, so we write $y=2 y_2$
We look at it $\mod 25$: this gives us
$7^y + 4 \equiv 49^{y_2} +4 \equiv (-1)^{y_2} + 4 \equiv 5^x \equiv 0$, which is impossible.
it remains to look at x=1, here we have the only solution $y=0,x=1,z=2$ (you can ignore this if you only want strictly positive solutions)
