a solution is here

I think it's number theory problem.

Would it be valid to invoke Mihailescu's Theorem? It wasn't a proven result in 2000, but it is now.

When I tried to see the solution, I found the link dead, so I post a new one: Clearly $y$ divides $x$, since $z\geq2$ hence $x^2$ divides $x^z=(x+1)^y-1=Sx^2+yx$, where $Sx^2$ is the sum of all terms in $(x+1)^y$ that have exponent $2$ or larger in $x$. Assume that $x>2$. Therefore, either $x$ has an odd prime factor $p$, or $x$ is divisible by $2$ with multiplicity at least $2$. Call $u\leq v$ the exponent of $p$ respectively in $x,y$ if $x$ has an odd prime factor, or $2\leqq\leqq$ the exponent of $2$ if $x$ has no odd prime factor. Now, we may write $$x^z=xy+\sum_{k=2}^y\binom{y}{k}x^k=xy+\sum_{k=2}\binom{y-1}{k-1}\frac{yx^k}{k}.$$ For any $k$, let $w$ be the multiplicity with which $p$ divides $k$. Clearly the numerator of $\frac{yx^k}{k}$ expressed as an irreducible fraction is divisible by $p$ with multiplicity $ku+v-w$, and since $\binom{y-1}{k-1}$ and $\binom{y-1}{k-1}\frac{yx^k}{k}$ are integers, then $\binom{y-1}{k-1}\frac{yx^k}{k}$ will be divisible by $p$ with multiplicity at least $ku+v-w$. Now, if $w\geq2$, then $k\geq p^w\geq2^w\geq w+2$, or $ku+v-w\geq k(u-1)+v+2\geq u+v+1$, because $k(u-1)+2\geq2(u-1)+2=2u\geq u+1$. If $w=1$ and $p$ is an odd prime, $k\geq3$ and $ku+v-w=u+v+(2u-1)>u+v+1$. If $w=1$ and $p=2$, then $k\geq2$, $u\geq2$, and $ku+v-w=u+v+(u-1)\geq u+v+1$ with equality iff $u=2$. Finally, if $w=0$, then $ku+v-w\geq2u+v\geq u+v+1$. Therefore, the sum on the RHS is divisible by $p$ with multiplicity at least $u+v+1$, but the first term is divisible with multiplicity $u+v$, hence the LHS is divisible by $p$ with multiplicity exactly equal to $u+v$, or $u+v=zu$. Now $y\geq2^v\geq v+1$ with equality iff $v=1$, hence $v\leq y-1$, and $zu=u+v\leq u+y-1$, ie $z=1+\frac{y-1}{u}\leq y$, or $1=(x+1)^y-x^z\geq(x+1)^y-x^y\geq yx^{y-1}\geq4$, absurd. We conclude that $x=2$. The equation then becomes $2^z=3^y-1=(3^t+1)(3^t-1)$, where $y=2t$ is even because $x$ is even and $x$ divides $y$. We therefore need two consecutive even numbers that are both powers of $2$, which means that $3^t-1=2$ and $3^t+1=4$, or $t=1$, $y=2$, and $2^z=3^2-1=8=2^3$, and the only possible solution is $(x,y,z)=(2,2,3)$.

This problem today is really simple considering lifting: $v_p(a)=\alpha$ implies that $p^\alpha \mid a$ and $p^{\alpha+1} \nmid a$ It's easy to see that $z>y$ $x^z=(x+1)^y-1^y \Rightarrow$ Since $(x,x+1)=1$ and $x\mid (x+1)-1$ Then $zv_p(x)=v_p(x^z)=v_p((x+1)^y-1^y)=v_p((x+1)-1) + v_p(y)$ $\Rightarrow v_p(y)=(z-1)v_p(x)$ If $4\mid x$ then $x^{z-1}\mid y$ $\Rightarrow z>y\ge x^{z-1}$ No solutions here. Then $v_2(x)=1 \Rightarrow z>y\ge (\frac{x}{2})^{z-1}\Rightarrow x=2$ and we solve the equation like @daniel73, so the only solution is $(x,y,z)=(2,2,3)$ $Q.E.D$

How did no one came up with this . Assume that $(x,z) \ne (2,3)$ then by Zsigmondy there exists a prime $p$ such that $p \mid x^z+1$ and $p \not \; \mid x+1$. Thus we have that $p \mid x^z+1=(x+1)^y$ thus $p \mid x+1$ giving a contradiction. Thus $x=2$ and $z=3$. By replacing we have: $3^y=9$ thus $y=2$ meaning that the unique sol is $(x,y,z)=(2,2,3)$ Thus we are done

Zsigmondy is useful!

Solution 1: This solution is really similar to @MathLuis's solution only because he hinted me at Zsigmondy's. We claim that the only solution is $(x, y, z)=(2, 2, 3)$. Now for the rest of the solution assume that $(x, z) \neq (2, 3)$. By Zsigmondy's, we have $p\mid x^z+1$. Since it $p$ is a primitive prime factor, we have $p\nmid x+1$. Now because $x^z+1=(x+1)^y$, we have $p\mid (x+1)^y$ which is a contradiction. Solution 2: Just use Catalan's conjecture and you are done.