TITLE: How many primes have the form $(2^p+1)/3$?
QUESTION [7 upvotes]: Assuming that $p$ is an odd prime.  How many primes have the form $(2^p+1)/3$? Is the number finite? Mathematica calculation shows that there are 23 such primes when $p$ ranges over the first 500 primes. Here are those primes $p$,
3, 5, 7, 11, 13, 17, 19, 23, 31, 43, 61, 79, 101, 127, 167, 191, 199, 
313, 347, 701, 1709, 2617, 3539 
Originally, I am concerned with prime powers of the form $(2^m+1)/3$, where $m>0$ is an odd number. This number can be a prime power only when $m$  is an odd prime. According to a result of T. N. Shorrey and R. Tijdeman (Math. Scand. 39, 5-18 (1976)), there are only finite number of such primer powers that are not primes, though I still don't know how many. Mathematica calculation seems to suggest there are no such prime powers that are not primes.

REPLY [8 votes]: Heuristically, the probability that a "random" number $X$ is prime is on the order of $1/\log(X)$, and the probability that it is the square of a prime is on the order of $2/(\sqrt{X} \log(X))$.  With $X = (2^m+1)/3$, that 
$$\dfrac{2}{\sqrt{X} \log(X)} \sim \dfrac{2\sqrt{3}}{m \log(2)} 2^{-m/2}$$
and the sum of this over $m$ converges quite quickly.  In particular the sum
from $m=1000$ to $\infty$ is approximately $5.2 \times 10^{-153}$.  Thus if none of these numbers
for $m < 1000$ are squares of primes, it seems quite likely that there are none at all.  Similarly for higher powers.  Of course this is just a heuristic rather than a proof, and should not be taken too seriously.