A typo-free detail oriented prime conjecture 
I've grossly overstated things in my two posts before the last one. Thank you for providing references that have returned me to reality.

Conjecture For arbitrary integers $\ 0 \le k \le m\ $ there exists
integer $\ n\ge m\ $ such that for every natural number $\ s\ $ at
least one of the numbers
$$\ p(x+s+1)-p(x+s)\ \ne\ p(x+1)-p(x) $$
where $\ k\le x < n$.

Here, $\ p(0)=2, p(1)=3,\ldots\ $ is the strictly increasing sequence of all prime numbers.

An assembler-like equivalent formulation:
Conjecture'
$$ \forall_{m\in\mathbb Z_{\ge 0}}\,\forall_{k\in 0..m}\,
\exists_{n\in\mathbb Z_{\ge m}}\,\forall_{s\in\mathbb N}\,
\exists_{x\in k..n\!-\!1} $$
$$ p(x+s+1)-p(x+s)\ \ne\ p(x+1)-p(x) $$

Here (Perl notation),

$$\ u..v\ :=\ \{x\in\mathbb Z:\ u\le x\le v\} $$
 A: Like in the previous post, let $q=p(k)$ and let $n\geq m$ be such that primes $p(x),k\leq x<n$ cover all residue classes mod $q$.
Suppose this $n$ doesn't work. This means that $p(x+s+1)-p(x+s)=p(x+1)-p(x)$ for all $k\leq x<n$. Adding up a bunch of such equalities we get $p(x+s)-p(k+s)=p(x)-p(k)$ for all $k\leq x\leq n$. As $s>0$, we have $p(k+s)>q$ so it is indivisible by $q$. There is some $k<x<n$ such that $p(x)\equiv -p(k+s)\pmod{q}$, and we get $p(x+s)=p(k+s)+p(x)-p(k)\equiv 0\pmod q$, which is impossible as $p(x+s)>q$, hence a contradiction.
A: 
NOTATION:
$u_0\ \ldots\ u_n\,\ $ and $\ v_0\ \ldots\ v_n\,\ $ are arbitrary finite strictly increasing sequences of non-negative integers, of the same sequence's length $\ n+1$.
$p(0)=2, p(1)=3, \ldots\ $ is the strictly increasing sequence of all
(consecutive) primes.
$ p(u_0)\ \ldots\ p(u_n)\,\ $ and $\,\ p(v_0)\ \ldots\ p(v_n)\ $ are
arbitrary finite strictly increasing sequences of primes (not necessarily
consecutive), where the length of the two sequences is the same, namely
$\ n+1.$
The goal: We will see that under certain additional assumptions, the
above two finite sequences are identical; in particular when one of them is assured to exist as an extension of a different shorter sequence.

Wojowu has posted his answer in the Q&A style. He deserves a more rounded presentation. Let me be the first one to do it. It'll be but a presentation. The results are exclusively due to Wojowu (Wojciech Wawrów).
Theorem 1 Assuming our NOTATION, if there exists a prime $\ q\ $
such that for every integer $\ x\ $ there exists at least one integer (index)
$\ a(x)\ $ such that
$$ p(u_{a(x)})\ \equiv x\ \mod q $$
and if there exists an integer $\ s\in\mathbb Z\ $ such that
$$\forall_{t=0}^n\quad p(v_t) = p(u_{s+t}) $$
and
$$ \forall_{t=1}^n\quad p(v_t)-p(v_{t-1})\ =\ p(u_t)-p(u_{t-1}) $$
then $\ s=0$.
Remark The last assumption can be written equivalently as
$$ \forall_{t=1}^n\quad p(v_t)-p(u_t)\ =\ p(v_{t-1})-p(u_{t-1}) $$
Proof (of Theorem 1) Let the assumptions of the theorem hold. By the above Remark,
for every integer $\ x\ $ there exists at least one integer (index)
$\ b(x)\ $ such that
$$ p(v_{b(x)})\ \equiv x\ \mod q $$
The terms in each prime sequence that are $\ 0 \mod q\ $
are simply equal to $\ q.\ $ If integer  $\ s\ $ were different from $\ 0\ $
then the same prime $\ q\ $ would appear twice -- $\ s\ $ places apart -- in the increasing sequence of all primes; that would be a contradiction.
End of Proof

More NOTATION: Given integers $\ 0\le m\le n,\ $ prime sequence
$ p(u_0)\ \ldots\ p(u_n)\,\ $ is called to be the minimal $n$-extension
of a strictly increasing sequence $ p(u_0)\ \ldots\ p(u_m)\,\ 
 \Leftarrow:\Rightarrow$
$$ \forall_{k=m+1}^n\quad u_k=u_m+k-m, $$

Theorem 2 For every strictly increasing finite prime sequence $ p(u_0)\ \ldots\ p(u_m)\,\ $ there exists a minimal extension $ p(u_0)\ \ldots\ p(u_n)\,\ $ which is lonely, meaning that it is the only prime sequence $ p(v_0)\ \ldots\ p(v_n)\,\ $ such that there exists an integer $\ s\in\mathbb Z\ $ such that
$$\forall_{t=0}^n\quad p(v_t) = p(u_{s+t}) $$
and
$$ \forall_{t=1}^n\quad p(v_t)-p(v_{t-1})\ =\ p(u_t)-p(u_{t-1}). $$
Proof For $n$ large enough all residue classes \mod p(u_0) appear
in $ p(u_0)\ \ldots\ p(u_n)\,\ $ (by Dirichlet Theorem). This, by Theorem 1,
makes sequence $ p(u_0)\ \ldots\ p(u_n)\,\ $ lonely. End of Proof
