Does there exist a maximum $m\in\mathbb{N}$ such that for some prime $p$, $p_r=p +\sum_{n=1}^r 2^n$ is prime for $r=1,2,\ldots,m$? Let $\mathbb{P}$ denote the set of prime numbers.

Does there exist a maximum $m\in\mathbb{N}$ such that for some prime $p$, $p_r=p + \sum\limits_{n=1}^r 2^n$ is prime for $r=1,2,\ldots,m$?

There is most likely not a simple (or known) answer to this question, but I thought it was worthwhile to ask.
So far, I have concluded that $p$ must necessarily have $7$ as its first digit, that is, $p=7\,(\text{mod}\,10)$. If $p=x\,(\text{mod}\,10)$ for some $x\in\{1,3,9\}$, the process is cut short at some $p_k=5\,(\text{mod}\,10)$.
For example, if $p=1\,(\text{mod}\,10)$, \begin{align} p_1&=3\,(\text{mod}\,10) \\ p_2&=7\,(\text{mod}\,10) \\ p_3&=5\,(\text{mod}\,10)\Rightarrow p_3\notin \mathbb{P} \end{align}
For $p=7\,(\text{mod}\,10)$, we have
\begin{align} p_{4k}&=7\,(\text{mod}\,10) \\ p_{4k+1}&=9\,(\text{mod}\,10) \\ p_{4k+2}&=3\,(\text{mod}\,10) \\ p_{4k+3}&=1\,(\text{mod}\,10)\end{align}
for $k\in\mathbb{N}\cup\{0\}$ (assume for simplicity that $p_0=p$).
Consider these two statements:
\begin{align} \exists p\in\mathbb{P}&:\;p_r=p + \sum_{n=1}^r 2^n\in\mathbb{P} \; \; \forall r\in\mathbb{N} \tag{1} \\ \exists m\in\mathbb{N}&:\;\text{for some $p\in\mathbb{P}$},\; p_r=p + \sum_{n=1}^r 2^n\in\mathbb{P} \; \; \forall r\in\{1,2,...,m\} \tag{2} \\ & \;\;\;\;\,\text{and}\; \forall p\in\mathbb{P},\; p_{m+1}\notin\mathbb{P} \end{align}
My strategy is simple: prove $(1)\Rightarrow (2)$ false OR prove $(2)\Rightarrow (1)$ false.
Through a bit of computerwork, I found that $m\ge 9$ for $p=2397347207$:
2397347207, 2397347209, 2397347213, 2397347221, 2397347237, 2397347269, 2397347333, 2397347461, 2397347717, 2397348229

If anyone can solve this, and/or point me toward helpful material, please post.
 A: A conjecture that the answer to your question is positive is very strong.  
Indeed, remark that $p_r=p + \sum\limits_{n=1}^r 2^n=p+2^{r+1}-2$ for each $r=1,2,\ldots,m$.

$\exists p\in\mathbb{P} :\;p_r=p + \sum_{n=1}^r 2^n\in\mathbb{P} \; \; \forall r\in\mathbb{N}\tag{1}$

If $\Bbb P$ is the set of prime numbers then (1) is false. Namely, $p_{p-1}=p+2^p-2$ is divisible by $p$ by Fermat little theorem. 
Thus the only way to have the positive answer is to have infinitely many distinct prime numbers $p$ such that both $p$ and $p+2$ are prime, that is they are prime twins.

So far, I have concluded that $p$ must necessarily have $7$ as its first digit, that is, $p=7\,(\text{mod}\,10)$.

Given $m$, we can obtain a family of similar restrictions on $p=p_0$ as follows. Let $\Bbb P_2=\{3, 5, 11, 13,\dots \}$ be the set of prime numbers $q$ such that $2$ has order $q-1$ in the (cyclic) multiplicative group $\Bbb Z^*_q$ of the finite field $\Bbb Z_q=\Bbb Z/q\Bbb Z$. Let $q\le m+2$ and $q<p$ be any element of $\Bbb P_2$. Then among the residues of $p_0,\dots, p_m$ we met all residues modulo $q$ but $p-2$. Since all $p_r$ are prime, this is possible only if the only missed residue is zero, that is if $p-2$ is divisible by $q$. That is $p-2$ is divisible by a product $\prod \{q\in\Bbb P_2: q\le m+2,\, q<p\}$.
