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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.

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    $\begingroup$ Observe that if $(2^{p}+1)/3 = q^{a}$ for some prime $q > 2$ and $a \geq 2$, then the $2^{2p} \equiv 1 \pmod{q^{a}}$ so the order of $2$ modulo $q^{a}$ is a divisor of $2p$, and also a divisor of $(q-1) q^{a-1}$, by Euler's theorem. Since $\gcd(p,q) = 1$, this implies that $2^{q-1} \equiv 1 \pmod{q^{a}}$, and so $q$ is a Wieferich prime. $\endgroup$
    – Jeremy Rouse
    Commented May 29, 2015 at 1:41

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I don't think anyone is going to answer this question, any more than anyone is going to decide the number of Mersenne primes any time soon, so for the sake of having an answer, I'll note that these numbers are tabulated at the Online Encyclopedia of Integer Sequences where they are called "Wagstaff numbers" and where many links and references are given.

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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.

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Such primes are called Wagstaff primes. Their (in)finiteness is an open question (similarly to Mersenne primes).

See this OEIS entry for further references: http://oeis.org/A000978

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