Integer solutions of $n^k+(n+1)^k+\cdots+(n+m)^k=(n+m+1)^k$ So we all know already that next identities follow:
$3^2+4^2=5^2$
$3^3+4^3+5^3=6^3$
So it raises my question:
For $(*)n^k+(n+1)^k+...+(n+m)^k=(n+m+1)^k$ are there infinite triples (n,m,k) s.t (*) is satisifed?
Any literature on this question or more general question from this follows?
Thanks.
 A: I do not have a definite answer, but in view of the discusion, and since it connects, as requested, the question to problems investigated in the literature, some remarks.
I claim that: 

If an answer to this question is known, then it is of the form that there are infinitely many solutions.

This might seem like a strange claim, and I do not claim that the number of solutions is infinite, but here is the reason: 
Erdős and Moser (in the 50s) conjectured that the equation 
$$ 1^k + \dots  + m^k = (m+1)^k $$
has only the solution $1+2=3$.
Over the decades it was investigated quite a bit, still it is unknown whether the number of solutions to this equation (the special case $ n= 1 $ of the questioner's equation) is even finite. 
So, even fixing $ n = 1 $ in the equation of the questioner one does not know how to prove that the number of solutions is finite.
Thus, the only way an answer to this question can be known, is the one I mentioned,
otherwise (and likely) this is an open problem.
I already mentioned the names Erdős and Moser; searching for 'Erdős--Moser type equation' will yield various investigations on this and related problems.
More specifically, the book 'Unsolved Problems in Number Theory' by Guy, already mentioned by Gerry Meyerson, contains a section on it (namely D7), you might be able to see it on Google books, with various references, in particular to computational work excluding solutions for $ m $ up to $ 10^{10^6} $ and even beyond that.
For more recent work on variants of this problem, see for example, a preprint by MacMillan and Sondow or a paper by Lengyel (scroll to A41). 
A: I'm quite unhappy with the answers I gave until now ; let me try to do better.
The first case you mention ($k=2$ and $m=1$) falls inside the theory of pythagorean triples ; that is the integer solutions to $x^2+y^2=z^2$. Some trivial calculations show that the only pythagorean triple which has the more specific form you ask for is the one you gave, with the relation $3^2+4^2=5^2$.
Then, the more general case you mention falls inside the search of integer solutions to an equation of the form $x_1^k+\dots+x_m^k=y^k$, where $k$ is fixed, so the equation is polynomial. It is for this general type of problem that I was mentioning that Faltings' theorem applies : if the geometric curve is too complex, then there will be only a finite number of solutions. And hence, only a finite number of solutions of the prescribed form.
I tried a few more calculations : for $m\geq2$ and $k=2$, I first saw that $n$ could only be $0$ or $1$ ; then using $\sum_{l=0}^Nl^2=\frac{N(N+1)(2N+1)}6$, I ruled out all possibilities.
For the other cases, I didn't find anything useful.
A: Can you find any other solutions aside from $1^1+2^1=3^1$ and $n^0=(n+1)^0$?  There certainly aren't any others with $k<100.$
You want $f(x,k,m)=x^k+(x+1)^k+...+(x+m)^k-(x+m+1)^k=0$ with $x$ a positive integer.  Let $x_{m,k}$ be the positive real root (if any). Convince yourself that, for $k \gt 2$, $2k-1 \lt x_{2,k} \lt 2k$. Next convince yourself that for fixed $m$, $x_{m,k}$ is a decreasing function of $k$ and that $x_{m,k} \lt 1$ if $m \gt \frac{3k}{2}$. This allows a quick search and the claim above.
A: "Introduction to elliptic curves and modular forms" by Neal Koblitz discusses at length the pythagorean triples (your first example)
More generally, your question falls in the arithmetic geometry realm, and there is a wealth of references. Let me just mention "Arithmetic of elliptic curves" by Joseph Silverman.
A: The question seems much harder than the pythagorean triples problem...
Now observe that:
$$n^k+(n+1)^k+...+(n+m)^k=1^k+2^k+...+(n+m)^k-(1^k+2^k+...+(n-1)^k)$$
So you can actually get an expression of this (using Bernoulli numbers). When $k$ and $n$ are fixed, it is polynomial in $m$ of degree $k+1$ so it equal to $(n+m+1)^k$ only for a finite number of $m$...
You could swap $m$ and $n$...
A: For the more general question, I think : Falting's theorem is what you want to know.
