All terms are primes. Largest currently known prime of the form (2^n + 1)/257 is (256^23029 + 1)/257 found by Donovan Johnson 03/2005. The only currently known prime of the form (2^n + 1)/65537 is (65536^239 + 1)/65537.
From Giuseppe Coppoletta, May 18 2017: (Start)
In general, for any j > 1, if (2^(n*2^j) + 1)/Fj is a prime (where Fj = 2^2^j + 1 is the corresponding Fermat number), then n needs to be prime, as for any odd proper factor q of n, 2^(q*2^j) + 1 is another factor of the numerator. The same for j = 0, apart for the particular value n = 3^2.
For the case j = 4, I checked it again, and (65536^p + 1)/65537 indeed is not a prime at least for 239 < p < 12500, i.e. (2^n + 1)/65537 is not a prime at least up to n = 200000. Any higher upper bound available?
One can also remark that 65536 = 2^16 and 239 = 2^8 - 2^4 - 1. Is there any special reason (see Brennen's link) for that?
I checked also that (2^(p*2^j) + 1)/Fj is never a proper power (in particular it is not a prime power) for j = 0..4 and for any prime p, at least for any exponent p*2^j < 200000.
We can even conjecture that ((Fj-1)^p + 1)/Fj is always squarefree for any odd prime p and for any Fermat number Fj with j >= 0. Note that this is not true if p is not restricted to be a prime, even if p and Fj are coprime, as shown by the following counterexample relative to the case j = 1, f1 = 5: 4^91 + 1 == 0 mod 1093^2. Remark that any such counterexample has to be a Wieferich prime (A001220), but not every Wieferich prime gives a counterexample, as shown by the second known Wieferich prime (3511), which cannot match here because it belongs to A072936.