I would like to know the solutions of next diophantine equation(s) involving the sequence of Stirling numbers of second kind ${r\brace s}$, for $1\leq s\leq r$. To create my variant I was inspired in last paragraph of the section D3 from [1].
Problem. Calculate the solutions $(m,n,z)$ of $${m\brace k}-{n\brace k}=z^k,\tag{1}$$ where $m\geq 1, n\geq 1$ and $z\geq 1$ are positive integers that satisfy the condition $$m>n,$$ and $k\geq 3$ is a fixed positive integer.
Question. For each integer $k\geq 3$ calculate all solutions of previous Problem. Justify your words. Many thanks.
I believe that previous problem isn't in the literature. Answer as a reference request if it is in the literature, adding the reference where is studied our diophantine equations and I try to search and read those propositions.
I suspect that the case $k=3$ of the Problem isn't similar to the case $k=2$ for which I know that has infinitelty many solutions. I'm not sure how to work the generalization that I've created as our Problem for the mentioned cases $k\geq 3$. I believe that it is possible to prove that the equation $(1)$ has no solutions (satisfying the required conditions) when $k$ is arbitrarly large, due an asymptotic approximation for Stirling numbers of the second kind.
I know that for the case $k=3$ that $(m,n,z)=(3,1,1)$ and $(3,2,1)$ are solutions. And for the case $k=4$ I know three solutions $(m,n,z)=(4,1,1),(4,2,1)$ and $(4,3,1)$. I don't know if it is obious to prove that the solution that we can glimpse are the only solutions (respectively for each case).
References:
[1] Richard K. Guy, Unsolved problems in number theory, Second Edition, Springer (1994).