(0,0,1,1) is a bonus solution. Notice even-odd possibilities, you see that x should be 0 . Then you get $3^y$+$5^z$= $7^u$-1 = $6p$ Look for mod 6 and you get y=0. Then $5^z$+2= $7^u$. Add -7 both. $5(5^{z-1}-1)$= $7(7^{u-1}-1)$. $5^{z-1}-1$≡0 (mod7) and (mod6). $7^{u-1}-1$≡0 (mod5) and (mod4) respectively. Probably there won't be any solution here but i couldn't solved yet.

Answer:$(x,y,z,u)=(0,0,1,1)$ When $x \geq 1$ $LHS$ is even, $RHS$ is odd so $x=0$. $3^y+5^z+1=7^u$ $(-1)^y+2 \equiv (-1)^u (mod 4)$ $y \geq 1 \implies (-1)^z+1 \equiv 1 (mod 3)$ Contradiction $y=0$. $5^z+2=7^u$ If $z \geq 2$ then $7^z \equiv 2 (mod 25)$ Contradiction $z=1 \implies u=1 \implies \boxed{(0,0,1,1)}$ $z=0 \implies 3=7^u$ No solution