The standard conjectures imply that there are infinitely many primes of the form $p=n^4+2$. No such prime can be part of a twin pair, since $p-2=n^4$ and $p+2=n^4+4=n^4+4n^2+4-4n^2=(n^2+2)^2-(2n)^2=(n^2+2n+2)(n^2-2n+2)$.
A simpler example is $p=n^2-2$.
An even simpler example is $p=21n+5$, and here we don't need any conjectures --- we know that there are infinitely many such primes $p$. Many more examples can be constructed along the same lines, e.g., $15n+7$, $15n+8$, $77n+9$, $39n+11$, etc., etc.
EDIT: The above was written before OP edited the question to indicate interest only in the five types of prime at the top of the question. So, let's look at those. In all cases, please ignore tiny counterexamples to generally true statements.
Mersenne primes $q=2^p-1$, $p$ prime. Trivially $q$ can't be the smaller of a pair of twin primes, so we are asking about $2^p-3$ and $2^p-1$ both being prime. Apparently this does happen from time to time, so there's no simple reason why it shouldn't happen infinitely often. On the other hand, we don't even know there are infinitely many Mersenne primes, so we're not going to prove it does happen infinitely often. In short: hopeless.
Sophie Germain primes: $p$ such that $p$ and $2p+1$ are both prime. $p-2$ is a multiple of 3, so we are asking whether there are infinitely many $p$ such that $p$, $p+2$, and $2p+1$ are all prime. The standard conjectures (e.g., Schinzel's Hypothesis H) say yes, but no one has a clue as to how to prove this. In short: hopeless.
Fermat primes. Maybe there are some congruences to show $2^{2^{2n}}+3$ can't be prime for sufficiently large $n$. It's worth a look. On the other hand, maybe there are only 5 Fermat primes anyway. There are heuristic arguments suggesting that there are only finitely many.
Regular primes. Another set that hasn't been proved infinite, although the smart money is leaning that way. I can't imagine any relation between the regularity of $p$ and the primality of $p\pm2$. Possibly just ignorance on my part, but I'm going to call this one: hopeless.
Fibonacci primes. The Fibonacci numbers grow exponentially, just like the powers of 2 (only not quite as fast), so the situation here is comparable to that with the Mersenne numbers. Hopeless.
Best Answer
Upon further thought, I realized that the answer must be no, we currently can't prove it.
If we could, a proof of the infinitude of polynomial primes immediately follows, since you can always split one into arbitrarily many parts. For example, the first six values of $x^2+1$ are $\{2,5,10,17,26,37\}$. But by substituting in $3x-2,3x-1,3x$ to $x^2+1$, it can be completely split into
each of which is itself admissible and contains a prime. To ensure admissibility, you either have to split into $k$ pieces where $k$ is not a residue of the original polynomial, or factor out $k$ if it is a residue, but aside from that, I see no reason why this wouldn't work in general, and since I'm sure I'm not the first person to have noticed this, it means that we can't prove that an admissible quadratic must have at least one prime.