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.
Observe that $5777\equiv 2\pmod 3$ and $2a^2\equiv 2\pmod 3$ unless $3\mid a$. Hence once you check that $5777-3$ is not twice a square, you need only check $a$ with $3\mid a$ (cutting down the effort by two thirds).
Likewise, $5777\equiv 2\pmod 5$, which allows you to drop all cases where $a\equiv \pm1\pmod 5$ (after checking that $5777-5$ is not twice a square).
Best Answer
This is because, in general, a conjecture is typically worded "Such-and-such is true for all values of [some variable]." So, a single counter-example disproves the "for all" part of a conjecture.
However, if someone refined the conjecture to "Such-and-such is true for all values of [some variable] except those of the form [something]." Then, this revised conjecture must be examined again and then can be shown true or false (or undecidable--I think).
For many problems, finding one counter-example makes the conjecture not interesting anymore; for others, it is worthwhile to check the revised conjecture. It just depends on the problem.