I did a search online and found a similar notion called Sophie Germain prime, which by definition is a prime $p$ such that $2p+1$ is also prime. Sophie Germain primes are conjectured to be of infinite many. I wonder if anyone has thought of the similar question replacing $2p+1$ by $2p-1$.
[Math] Are there infinite many primes p such that 2p-1 is also prime
number theoryprime numbers
Related Solutions
Claim : $$2^{2^n+1}+3$$ is divisble by $5$ for $n\ge 2$
Proof : Because of $2^4\equiv 1\mod 5$ we can reduce the exponent modulo $4$, which gives $2^1+3=5$ which is divisble by $5$. QED
Hence there is no further prime of the form $2F_n+1$ and therefore no further Fermat-Sophie-Germain-prime.
Check this out: https://en.m.wikipedia.org/wiki/Sophie_Germain_prime
What you've described is apparently called a Cunningham chain. It has been conjectured that they can be made arbitrarily long, but it cannot be infinitely long.
In terms of your conjecture, it is conjectured that there exists some prime(s) that have a specified depth $d$, but no prime has an infinite depth.
I think I have a proof that all sequences $x_n$ where $x_n= ax_{n-1} +b$ can't be prime for all $n$, but I'll have to do it later.
Not sure if you've found a proof for the result above (or care still), but I found where I had proven it before. Since it's kinda long, I'll paraphrase it a bit, although if you need clarification feel free to comment.
Denote $(a,b)$ by the greatest common divisor between $a$ and $b$. For example, $(45, 30) = 15$. In particular, if $(a,b) = 1$, we say that $a$ and $b$ are relatively prime.
Start by writing $x_n = x_0 + b(n-1)$ for $a = 1$ and
$$x_n = a^nx_0 + b\left(\frac{a^n-1}{a-1}\right)$$
for $a >1$.
Note that $x_n$ is always divisible by $(a,b)$ and $(b,x_0)$. Hence, we will assume $(a,b) = (b,x_0) = 1$.
Showing the statement for $a=1$ is simple by considering $x_{n+1}$ modulo $k$ where $(k,b) = (k,x_0) = 1$. It then follows that
$$n \equiv -x_0b^{-1} \pmod{k}$$
makes $x_{n+1}$ divisible by $k$.
Now, for $a > 1$, take some $k$ so that $k$ is relatively prime to $a,b,x_0$ and define a function $n(t)$ over positive integers $t$ by
$$n(t) = t\phi(k) + 1,$$
where $\phi$ is the Euler-totient function.
Consider $x_{n(t)}$ modulo $k$,
$$x_{n(t)} = a^{t\phi(k) + 1}x_0 + b\frac{a^{t\phi(k) + 1}-1}{a-1} \equiv ax_0 + b \pmod{k},$$
where we used the Euler-Fermat theorem. Since $(a,b) = (b,x_0) = 1$, $ax_0 + b = x_1$ is relatively prime to each of $a,b,x_0$.
Thus, $k = x_1$ satisfies the conditions of $k$ specified, so taking $k = x_1$ implies that each member of the sequence $\{x_{n(t)}\}$ is divisible by $x_1$. But, $\{x_{n(t)}\}$ is strictly increasing, so only $x_{n(1)}$ can possibly be prime.
Therefore, for $t > 1$, $x_{n(t)}$ is composite.
Best Answer
These primes are tabulated at http://oeis.org/A005382
They don't seem to have a name, though they are related to Cunningham chains.