Suppose for contradiction that all terms of the progression $(a_k)$ are prime numbers.

Let us consider the two possible recursions:

- $a_k = 2a_{k-1} + 1$
- $a_k = 2a_{k-1} - 1$

Let $a_1 = n > 1$.

Let us compute $a_2$:
- $a_2 = 2n + 1$ or $a_2 = 2n - 1$

Let us consider the sequence where we always choose $a_k = 2a_{k-1} + 1$.

Then $a_2 = 2n + 1$, $a_3 = 2(2n + 1) + 1 = 4n + 3$, $a_4 = 2(4n + 3) + 1 = 8n + 7$, and so on.

In general, if we always choose $a_k = 2a_{k-1} + 1$, then $a_k = 2^k n + (2^k - 1)$.

Similarly, if we always choose $a_k = 2a_{k-1} - 1$, then $a_k = 2^k n - (2^k - 1)$.

Let us consider $a_k$ modulo $a_1 = n$.

For the sequence $a_k = 2^k n + (2^k - 1)$:

$a_k \equiv 2^k n + (2^k - 1) \pmod{n}$

$\equiv (2^k - 1) \pmod{n}$

So for $k = n$, $a_n \equiv (2^n - 1) \pmod{n}$.

If $n$ is odd and greater than $1$, then $2^n - 1$ is even, so $a_n$ is divisible by $n$ for some $k$.

But more generally, for large enough $k$, $a_k$ will be divisible by $n$ or by some smaller $a_j$.

Alternatively, note that for any $n > 1$, the sequence grows rapidly, and for large enough $k$, $a_k$ will be composite.

For example, if $n = 2$, then $a_1 = 2$.
- $a_2 = 5$ or $3$
- $a_3 = 11$ or $5$
- $a_4 = 23$ or $9$

If we choose $a_2 = 3$, then $a_3 = 5$, $a_4 = 9$ (which is not prime).

If we choose $a_2 = 5$, $a_3 = 11$, $a_4 = 23$, $a_5 = 47$, $a_6 = 95$, which is not prime.

Therefore, for any starting $n > 1$, not all terms of the progression can be prime numbers.

Thus, not all terms of this progression are prime numbers.