How many prime with a power exist between $n^2$ and $(n+1)^2$? How many prime with a power exist between $n^2$ and $(n+1)^2$ ?
For example if $n=5$ , there exists 27=3^3 between $25$ and $36$.
I'm searching for an upper bound for the number of primes with a power
existing between $n^2$ and $(n+1)^2$.
The prime is less than or equal to $n$ and the power is more than 1
 A: In the range $(n^2, (n+1)^2)$ there is at most one one 3rd power, one 5th power, and 7th power, etc. The largest power which is possible is $2^{\log_2(n^2+2n)}$ and so an upper bound is $\pi(\log_2(n^2+2n))-1.$ So for $n=100$ there are at most 5 proper prime powers and for $n\le10^{100}$ there are at most 120 proper prime powers.
I should mention that I can't find any such intervals with more than 2 proper prime powers, and that heuristics suggest that there are only finitely many with more than two. $n=46$ is the last example I can find with two in the interval, having both $13^3$ and $3^7$. Others must have $n>10^9.$ (Edit: A117934 shows that $n>7.9\cdot10^{23}$.)
However, heuristics suggest (perhaps counter-intuitively?) that there should be infinitely many intervals with two nontrivial prime powers. In particular the 'chance' that there is another with $n<10^{18}$  is around 50/50.

Actually, I can improve the bound further. Note that since there are no primes between 2 and 3, $\log_2(n^2+2n)$ can be replaced by $1+\log_3(n^2+2n)$. Similarly this can be replaced by $2+\log_5(n^2+2n)$, $3+\log_7(n^2+2n)$, etc., all of which grow more slowly than the last. In the limit it makes sense to use about $\log n/\log\log n$ primes, making the argument $O(\log n/\log\log n)$ and hence the bound 
$$
O\left(\frac{\log n}{(\log\log n)^2}\right).
$$
Applying this technique to the numbers above, you can see that there are at most 4 nontrivial prime powers in intervals with $n\le100$ (using $1+\log_3$) and at most 40 in intervals with $n\le10^{100}$ (using $9+\log_{29}$).
A: This is very heuristically: 
Consider the intervals $[n^2,(n+1)^2)$ for $n=1,2,\ldots, N$. Together these $N$ intervals of interest cover $[1,(N+1)^ 2)$, a range n which there are $\pi(N+1)\approx\frac N{\ln N}$ prime squares, $\pi(\sqrt[3]{(N+1)^2}\approx \frac{N^{\frac23}}{\frac32\ln N}$ prime cubes and so on. The first expression is dominant, and as we ultimately exclude squares, the next biggest summand (the one for cubes) becomes dominant. So essentially there are $\approx \frac {N^{\frac23}}{\frac23\ln N}$ prime powers in the first $N$ intervals of interest, hence on average only $\frac{3}{2\sqrt[3]N\ln N}\ll 1$ in each. This is also a rough estimate for the number of prime powers in the last interval $(N^2,(N+1)^2)$.
Continuing with handwavy heuristics, we may think of the number of prime powers in the $N$th interval as being Poisson distributed with mean $\frac{3}{2\sqrt[3]N\ln N}$ and that makes (for $N$ large enough) the probability of two prime powers in the intrval $\approx \frac12\left(
\frac{3}{2\sqrt[3]N\ln N}\right)^2$, which is even smaller. Nevertheless, summing this expression over $N$ diverges, thus hinting that there should be infinitely many cases where two prime powers occur. The same argument suggests that infinitely many cases with three prime powers occur, but for four powers we arrive at a converging series! 
Hence it is feasible that  more than three prime powers occur only finitely often. Taking Charles' findings into account, one might dare to conjecture that there is a finite bound and that it may be as small as $4$ (or even $2$).
