Let $x+1 = a$, $y+1=b$ and $z+1 =c$ Now the equation becomes $a^b + 1 = (a+1)^c$ and $a$,$b$,$c$ are positive integers. Firstly for $c=1$ or $b=1$ we get that the equation is satisfied for all natural $a$ and $b=1$ or $c=1$ respectively. So let $c \ge 2$ and $b \ge 2$ Taking mod $(a+1)$, we get that $b$ is odd. Therefore $a^b +1 = (a+1)(a^{b-1} - a^{b-2} + ..... +1) = (a+1)^c$ Again taking mod $(a+1)$ and since $c \ge 2$, we get $a+1 | b$ Since $b$ is odd, $a+1$ is also odd. So $a$ is even. Now considering the binomial expansion of $(a+1)^c$, we get we have $a | c$ So $c$ is also even. Let $a=2m$, $c=2n$ So we get, $2^{m}b^{m} = [(a+1)^n + 1][(a+1)^n - 1]$ Since $gcd[(a+1)^n + 1],[(a+1)^n - 1] = 2$ and since $2m | (a+1)^n -1$, $ \Longrightarrow (a+1)^n -1 = 2m^b$ and also $(a+1)^n + 1 = 2^{b-1}$ $\Longrightarrow 2^{b-1} > 2m^{b} \Longrightarrow m=1$ So all solutions are $(x,y,z) = (x,0,0) , (1,2,1)$

dibyo_99 wrote: Let $x+1 = a$, $y+1=b$ and $z+1 =c$ Now the equation becomes $a^b + 1 = a^c$ and $a$,$b$,$c$ are positive integers. It should be $a^b + 1 = (a+1)^c$.

Sorry, typo. Edited.

We can reduce the problem to $a^b+1=(a+1)^c$ We may use the catalan's conjecture which was proved by Mihilescu(also known as Mihilescu's theorem) which states that $x^y-z^t=1(x,y,z,t >1)$ has only one solution in integers that is $x=t=3,y=z=2$ Now in the above case we get $a=c=2,b=3$ or otherwise at least one of $a,b,c$ must be $1$ $a=1 \Longrightarrow (a,b,c)=(1,b,1)$ $b=1 \Longrightarrow (a,b,c)=(a,1,1)$ $c=1 \Longrightarrow (a,b,c)=(a,1,1) \Longrightarrow (x,y,z)=(1,2,1),(0,m,0),(m,0,0)$

That may be too much for this problem. But using Zsigmondy's theorem, there will be a prime factor of $a^b+1$ which won't divide $a+1$ gives us a contradiction. Otherwise, we can use the following simple lemma: https://www.awesomemath.org/assets/PDFs/Exponent_GCD_Lemma.pdf

Just a Mihailescu's theorem

Lol @aboves. Mihăilescu's theorem was given in 2002, and this is a TST from 2000 First solution misses $(0, y, 0)$ - from the second solution (post #5), but this uses Mihăilescu's ... is there a way to save this first solution?

Another French Diophantine which is trivial by zsigmondy :p Case 1: $y=0$ hence we get that $(x,y,z)=(x,0,0)$ works and its trival that is the unique here. Case 2: $y>0$ Assume that $x \ne 1$ and $y \ne 2$ then by zsigmondy there exists $p$ such that $p \mid (x+1)^{y+1}+1$ but $p \not \; \mid x+2$ which is not possible, hence $x=1$ and $y=2$ and this tells that $z=1$ hence $(x,y,z)=(1,2,1)$ Thus we are done