TITLE: Solve $(x+1)^y-x^z=1$
QUESTION [6 upvotes]: I want to know if there is another way to solve the equation:
$$(x+1)^y-x^z=1$$
for positive integers $x,y,z$ greater than $1$ without using the Catalan identity. Does exist? By now I only know that $x\mid y$.

REPLY [2 votes]: In Which is greater, 9899 or 9998? it was stablished that $(x+1)^x<x^{x+1}$ for all real $x>2.293$. On the other hand, $(x,y,z)=(1,1,1)$ ,$(2,2,3)$ and $(1,1,z)$ with $z\ge1$ are clearly solutions.
We have  $$(x+1)^y-x^z=1 \Rightarrow (x+1)^y>x^z\Rightarrow y<z$$ From
$$\begin{cases}(x+1)^x<x^{x+1}\text{ for all x>2.293}\\(x+1)^y>x^z\\y<z\end{cases}$$ we deduce easily $y<2.293$ then $y=1$ or $2$ which gives the only solutions above.