Can we find prime in the form of $2^nx+1$ for arbitrary $x>0$?

Question

For example, is it so that if we append enough number of $0$s followed by a $1$, any element $x\in \mathbb{Z}_p^*$ can be interpreted as a prime number, where $p$ is a $k$-bit prime for parameter $k$? (e.g. $x0000\ldots 1$)

[update]

What I mean is that the $0$s and $1$ are binary bits. So suppose $p=2^{127}-1$, $x=2^{126} \iff x=\underbrace{1000\ldots00_2}_{126}$. Can we append, say $n$ $0$s and a $1$ such that $x'=x\underbrace{00\ldots0}_{n}1_2=\underbrace{1000\ldots00_2}_{126+n}1$ is a prime?

And can we do so for arbitrary $x\in \mathbb{Z}_p^*$?

[Update]

Let's ignore the $p$ and $k$, they aren't relevant in general. What Chris Culter said is right: can we find prime in the form of $2^nx+1$ for arbitrary $x>0$?

Answer 1

$$ 78557 \cdot 2^n+1 $$ is always composite (as proved by Selfridge via covering congruences). Finding such numbers is an old problem of Sierpinski.

Answer 2

If interpreted as a standard decimal integer, the answer is no, because 2 followed by any number of zeros and then a 1 is divisible by 3.

Similarly, in any base B, the digit B-2 followed by any number of zeros and a 1 is divisible by B-1.