Consider by mod (8) give $x=y=z(mod \ 2)$. Obviosly solutions $x=y=z=0$ and $x=y=z=1$. I think there are not another solutions.

I don't understand you, @Rust. Who can help me?

Let $y=0.$ Then $7^x=5^z\to x=z=0$. Let $y=1$. Then $7^x=5^z+2.$ Therefore $z\ge 1, x=1\mod 4$ (consider by mod 5). If $z=1$, then $x=1$. If $z>1$, then $7^4=1\mod 25$, therefore $7^x=7\mod 25$ and $ 7\not =2\mod 25.$ There are not solutions, suth that $z>1$. Now consider $y>1$. Must be $7^x+1=3^y+5^z\mod 9, x=1\mod 4$. Because $7^x+1=2\mod 3\to z -odd \to z=3\mod 6$. Consider by mod 7 $1=3^y-1\mod 7\to y=2\mod 6$. If $y=2\mod 6, z=3\mod 6$ $3^y+5^z=6\mod 8$. But never $7^x+1=(-1)^x+1\not =6\mod 8$. Therefore only $x=y=z=0$ and $x=y=z=1$ are solutions.

We wil show that the only solutions are $(0,0,0)$ and $(1,1,1)$. I can follow the proof of Rust up to the point where he concludes that z=3 mod 6, which I don't follow. I think he wants to use $x\equiv 1(4)\Rightarrow 7^x\equiv 7(9)$ but that is not true, because the order of 7 modulo 9 equals 3 and not 4. Hence we will follow a slightly different method. As in his proof another solution has $x,y,z>0$ and $y>1$, which we assume. Modulo 3 one has $2\equiv 5^z$, hence $z$ is odd. Modulo 4 one has $(-1)^x+1\equiv (-1)^y+1$, hence $x\equiv y(2)$. Modulo 5 one has $2^x+1\equiv(-2)^y$. If $x,y$ are even with $x=2a,y=2b$, we have $(-1)^a+1\equiv(-1)^b$. Since $\{0,2\}\cap \{-1,1\}=\emptyset$ this is a contradiction, hence $x,y$ are odd. Reading again $2^x+1=3^y$ and noting that $(\{2,-2\}+1)\cap \{3,-3\}=\{3\}$ , we find $x\equiv y\equiv 1(4)$. Modulo 16 we have $7^2\equiv 1,3^4\equiv 1$, hence $7^x\equiv 7,3^y\equiv 3\Rightarrow 5^z\equiv 5\Rightarrow z\equiv 1(4)$. Hence $x,y,z\equiv 1(4)$. We now combine $3^y+5^z\equiv 1(7)$ with $7^x+1\equiv 5^z(9)$ (here we use $y>1$) and consider the possible residues of $x\mod 3$. If $x\equiv 0(3)$ then $7^x\equiv 1(9)\Rightarrow 5^z\equiv 2(9)\iff z\equiv 5(6)\Rightarrow 5^z\equiv 3(7)\Rightarrow 3^y\equiv 5(7)\iff y\equiv 5(6)$, hence $(x,y,z)\equiv (9,5,5)\mod 12$. But then we can read $7^x+1=3^y+5^z$ modulo 13 and obtain $7^9+1\equiv 3^5+5^5$, not true. The other cases, $x\equiv 1,2(3)$ are similar and left to the reader.

N.T.TUAN wrote: Find all $ x,y,z\in\mathbb{N}_{0}$ such that $ 7^{x} + 1 = 3^{y} + 5^{z}$. Alternative formulation: Solve the equation $ 1+7^{x}=3^{y}+5^{z}$ in nonnegative integers $ x$, $ y$, $ z$. With mod$8$ we get that $x=y=z(mod2)$ If $y=0$ esily we get $x=z=0$ If $y>=1$ with mod$3$ we get $x,y,z$ are all odd If$y=1$ we get $7^x=5^z+2$ with mod $5,25$ we get the only sollution $x=y=z=1$ If $z=1$ we get $7^x=3^y+4$ with mod$16$ we get that $y=1(mod4)$ with $mod7$ we get that $y=1(mod6)$ so $y=1(mod12)$ with $mod13$ we get that $x=1(mod12)$ with $mod9$ if $y>1$ we get a contradiction. Suppose now that $y,z>1$ then we have: With $mod9$ we get $(x,z)=(1,3),(2,1),(0,5)(mod (3,6))$ With $mod5$ we get $(x,y)=(1,1)(mod (4,4))$ With $mod7$ we get $(y,z)=(1,1)(5,5)(mod(6,6))$ So we have that $(x,y,z)=(5,1,1),(9,5,5)(mod(12,12,6))$ which givew contradiction $mod13$ So the only sollution are $(x,y,z)=(0,0,0),(1,1,1)$