Twin primes of form $2^n+3$ and $2^n+5$ How to prove that $2^n+3$ and $2^n+5$ are both prime for only finitely many integers $n$?
And how to prove that there are infinitely many primes of the form $2^n+3$ and $2^m+5$
 A: 
And how to prove that there are infinitely many primes of the form $2^n+3$ 

First prove that there are infinitely Mersenne primes, then...
This is something that can be predicted, not proved, in the current state of number theory.
The "probability" of prime numbers near $2^n$ is close to
$1/\log (2^n) = \frac{1}{n \log 2}$ and the "expected number" of primes of the form $2^m+3$ in $[1,n]$ is $O(\log \log n)$.
A: What makes you think that there are only finitely many?  A057732 shows a large number of primes of the form $2^n+3$, and likewise A059242 for $2^n+5.$
Generally the way that such sequences are shown to have only finitely many primes is to find a covering set, but there's no obvious one for either.

To answer the edited question: There are only finitely many twin primes of the form $(2^n+3,2^n+5)$ because there is a covering set: at least one of the numbers is divisible by a member of $\{3,5,7,13\}$ (work mod 12).  But proving that there are infinitely many primes of either form is beyond current technology AFAIK.
A: Settling the twin prime part of the question seems to go as follows:
1) If $2\mid n$, then $3\mid 2^n+5$, so it suffices to study odd values of $n$.
2) If $n\equiv 1\pmod 4$, then $5\mid 2^n+3$, so it suffices to study the case $n\equiv 3\pmod 4$.
3) If $n\equiv 1\pmod 3$, then $7\mid 2^n+5$. If $n\equiv 2\pmod 3$, then $7\mid 2^n+3$, so for there to be infinitely many twin primes of this form we must have $3\mid n$.
Combinining items 2 and 3 above, we see that it suffices to exclude the possibility of infinitely many twin primes of this form, when $n\equiv 3\pmod {12}$. But...
4) If $n\equiv 3\pmod {12}$, then $13\mid 2^n+5$.
Looks like $(5,7)$ and $(11,13)$ are all.
