Disprove the Twin Prime Conjecture for Exotic Primes The List of unsolved problems in
mathematics
contains varies conjectures of exotic primes like:

*

*Mersenne primes (of the form $2^p - 1$ where $p$ is a prime,
A000668, $43\%$)

*Sophie Germain primes ($p$ and also $2p+1$ is prime,
A005384, $42\%$)

*Fermat primes (of the form $2^{2^k} + 1$, A019434, $100\%$)

*regular primes
(A007703, $61\%$)

*Fibonacci primes
(A005478, $44\%$)

to name just a few. Each of these primes is provided with special
property. Sitting enthroned over the list on the wiki page you'll find the Twin
Prime Conjecture (ignoring Catalan's Mersenne conjecture for the
moment). So here is my

Question: Assuming the infinitude of all exotic primes, for which is it possible to disprove the infinitude of twin prime pairs containing one exotic prime?


Example
It could be done for e.g. primes of the form $p_n=(6n)^2+1$, where
it's obvious to show that $p_n\pm 2$ are composite and $p_n$ is called
isolated prime.
And the weaker $5$th Hardy-Littlewood conjecture asserts that $a^2+$1
is prime for an infinite number of integers $a>1$ [from Bouniakowsky
Conjecture].
Don't get me wrong: This is not what I'm looking for! The question is on the primes given in the list above: Fermat/SophieGermain/Mersenne/... and how to disprove that they are infintely often one of the twins in a pair. Mea culpa, if this is misleading.
Collected partial results

*

*Percentages given in the list above, show the ratio of exotic primes having
a twin (some where counted twice, specially in the regular case).


*For Fermat primes it seems promising ($100\% !$) to prove that every prime has a twin, but then this restricts to Fermat primes of the form $2^{2^{2n}}+1$, since
$7 \mid 2^{2^{2n+1}}+3$ (see coment below) and $2^{2^{2n+1}}-1$ is obviously composite.


*An analogous analysis might done for Mersenne primes, but I haven't yet.


*The Wiki page on Twin
Primes gives some more
general ways to tackle the problem, but I'm not sure, if they are really useful:



*

*Every twin prime pair except $(3, 5)$ is of the form $(6n - 1, 6n +
1)$ for some $n$, and with $n \neq 1$, $n$ must end in $0, 2, 3, 5, 7
\text{ or } 8$. -- This seems related to my example given above, since $1,4,6$ and
$9$ are missing, which show up as end digits of square numbers, see
here. $0$ and $5$ seem
to be exceptions.


*The pair $(m, m+2)$ is twin prime, iff $4((m-1)! + 1) \equiv -m \pmod
{m(m+2)}$.
So if you feel that you can disprove the twin prime conjecture on any of these exotic primes, I'd be every so happy to read your answer here. If you think you can prove it for a kind of primes, where the infinitude is also proven, send me an eMail.
 A: There are almost surely infinitely many regular primes which are part of a twin prime pair. The first few are

3, 5, 7, 11, 13, 17, 19, 29, 31, 41, 43, 61, 71, 73, 107, 109, 137, 139, 151, 179, 181, 191, 193, 197, 199, 227, 229, 239, 241, 269, 281, 313, 349, 419, 431, 521

and there are 1513 below $10^5.$
I disagree with Gerry on Mersenne primes. While we expect that there are infinitely many, there should be only finitely many which are part of a twin prime pair since you'd need $2^p-3$ and $2^p-1$ to both be prime and this happens with probability $\sim k/p^2$ and $\int dx/x^2$ converges.
For Sophie Germain primes, this would follow from Dickson's conjecture and so this one is perhaps the closest of all these questions to actual resolution. This is related to A045536 but sadly no information is present there.
Fermat primes are not expected to be infinite in number, so there should not be infinitely many twins.
Fibonacci primes are spread pretty thin, so like with the Mersenne primes I expect there are only finitely many which are twins.
A: In the case of Fibonacci twins and Mersenne twins, we should strongly expect those sets to be finite, although proving it will be difficult. Here's the essential heuristic:
A standard approach to asking if a set of primes is infinite or finite is to act like it is a random set of numbers where the probability of a number being in the set is $1/\log n$. This is essentially using the prime number theorem which can be thought of as saying that a number $n$ is prime with probability about $1/\log n$. 
For example, for the Mersenne primes, we believe that there are infinitely many in part because we can take say that $1/\ln(2^p-1)$ is about a constant times $p$ and so $$\sum_{p} \frac{1}{\ln(2^p-1)}  \approx \sum_{p} \frac{1}{\ln(2^p)}  \approx \sum_{p} \frac{1}{p\ln(2)} \approx \infty . $$ Essentially this means that if we go out far enough, no matter how far we go, the "probability" that there's another Mersenne prime should be very high (note that we're using in the above that the sum of the reciprocals of the primes diverges). But if we try to do the same sort of approach for your twin Mersenne primes we get
$$\sum_{p} \frac{1}{\ln(2^p-1)}\frac{1}{\ln(2^p-3)}  \approx \sum_{p} \frac{1}{\ln(2^p)}\frac{1}{\ln(2^p)} \approx  \sum_{p} \left(\frac{1}{p\ln(2)}\right)^2 \ll \sum_{n=1}^\infty \frac{1}{n^2} < \infty . $$ So if we go out far enough, the probability of finding such a pair ends up getting very small. (This is the same sort of heuristic approach that leads us to expect only finitely many Fermat primes.)
Similar logic works for the case of a Fibonnaci twin pair, and working out the details of the heuristic may  be a useful exercise. 
Regarding the Fibonnaci twin primes, note that there's another possibly interesting variant of this problem. Let $F_n$ be the $n$th Fibonacci number. Then if $F_p$ is prime then one must have $p$ is prime; this is a well known result which follows from the fact that $m|n$ then $F_m|F_n$. In a paper by Sean Bibby, Pieter Vyncke, and me currently under review (arXiv version here), we had occasion to ask if there are infinitely many $n$ such that $F_n$ and $F_{n+2}$ are both prime. Any such pair requires that $n$ and $n+2$ are a twin prime pair, and a similar heuristic would suggest that this set if likewise finite (see page 28 of that preprint).  This problem is also probably hard. 
A: 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. 
