I have not seen this question treated in the literature. Does anyone have more information? There are several OEIS sequences (A097056, A117896, A117934) dealing with this question, but no answers.
If you let $a_1,a_2,\dots$ represent the sequence $1,4,8,9,\dots$ of perfect powers, it is a conjecture of Erdos that $$a_{n+1}a_n >c'n^c.$$ The weaker conjecture $$\liminf a_{n+1}a_n =\infty$$ is known as Pillai's conjecture. This seems to be much beyond the reach of the strongest methods available today. Note, however, that your conjecture in the OP implies $$a_{n+2}a_n>cn^{1/2}$$ which is a very strong, related result. To give an idea of how hard it is to prove that consecutive powers tend to space out more and more, consider how hard it was to prove Catalan's conjecture (consecutive powers can not be 1 apart, except for 8 and 9) or even the still open Hall's conjecture which asserts that if $k=x^3y^2\neq 0$ then $$k>\max\{x^3,y^2\}^{1/6}o(1)$$ which is related to some abc type conjectures. Now on the specific problem you mention, there is some work done on powers in relatively short intervals. Start with
where it is proved that the interval $[N,N+\sqrt{N}]$ contains at most $$\exp(40(\log\log N \log\log\log N)^{1/2})$$ integer powers. This is far from the bound of $3$ but it is a nice result. There was a gap in the paper above, but it was corrected by two different authors, see
and
In the first paper by Stewart above, it is conjectured that there are infinitely many integers $N$ for which the interval $[N,N +\sqrt{N}]$ contains three integers one of which is a square, one a cube and one a fifth power. He also conjectures that for sufficiently large $N$ this interval does not contain four distinct powers and if it contains three distinct powers then one of is a square, one is a cube and the third is a fifth power, which is a refined version of your conjecture. For more references and a longer survey see Waldschmidt's "Perfect Powers: Pillai’s works and their developments". 


It follows from the ABC conjecture that there are only finitely many. First, show (unconditionally) that for all but finitely many such triples of perfect powers, at most one of them is a cube. So, the others are fifth or higher powers, say $n^2<A<B<(n+1)^2$. Now, apply ABC to $BA=C$. Then $B>n^2, C = O(n)$ and the radicals of $A$ and $B$ are $O(n^{2/5})$, so the ABC conjecture bounds $n$. 


From http://oeis.org/A097056/internal: is probably the closest you'll ever get to an answer. 

