How many primes have the form $(2^p+1)/3$?

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.

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.

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