Can we find $n$ such that $3061\cdot2^n +1$ is prime? Let $p=3061$. Can we find an integer $n$ such that $2^n p+1$ is prime? If there is no such $n$, how can we prove it? Or does such $n$ always exist for prime $p$?
(More generally, instead of $p=3061$, you can try e.g. $p=5297,5897,7013,8423,\ldots$ -- there are quite a few primes $p$ for which brute force does not seem to work.)
Motivation: questions like these arise naturally while reading the paper 
 On the density of odd integers of the form $(p − 1)2^{-n}$ and related questions
by Paul Erdös and Andrew Odlyzko.
 A: Not a complete solution but some observations. Let $a_n=2^np+1$.
I.  If $n$ is odd then $3\,|\,a_n$.
Pf:  Indeed  $3061\equiv 1 \pmod 3$ so $a_n\equiv 2^n+1\pmod 3$ and the claim follows.
II.  If $n\equiv 2 \pmod 4$ then $5\,|\,a_n$
Pf:  Indeed, $3061\equiv 1 \pmod 5$ so $a_n\equiv 2^n+1\pmod 5$ and the claim follows.
So you only need to worry about $n\equiv 0 \pmod 4$.  For those there does not appear to be a simple congruence.  Indeed, for $n\in \{1,\cdots 10\}$ the least prime which divides $a_{4n}$ is, respectively, $$\{17,19,17,797, 17,821,17,31,17,59\cdots\}$$
From which we are led to conjecture that $n\equiv 4 \mod 8\implies 17\,|\,a_n$, which is easily proven (since $3061\equiv 1\pmod {17}$ and $2^4\equiv -1 \pmod {17}$). 
That just leaves $n\equiv 0\pmod 8$ to study.
A: These are Proth numbers so you can apply Proth's theorem for efficient primality testing (see also Are there primes $p=47\cdot 2^n+1$?.). This way, we can verify that the pseudoprime posted by @DmitryEzhov in comments is indeed a prime by choosing $a=3$, $N=3061\cdot 2^{33288}+1$, because
$$
a^{\frac{N-1}{2}} \equiv -1 \pmod{N}.
$$
(e.g. a := 3: p := 3061*2^33288+1: is((Power(a, (p-1)/2) mod p) = p-1) in Maple).
Also Effective Primality Test for $p2^n+1$, $p$ prime, $n>1$ could be of interest. It especially mentions:

Sierpinski numbers of the second kind are integers $k$ such that $k2^n + 1$ is not prime for all positive integers $n$.

So in your case, if the Sierpinski number $k=p$ happens to be a prime, you will not find $n$ such that $p2^n+1$ is a prime. There are infinitely many such numbers, the  $k=271129$ is conjectured to be the smallest (see Prime Sierpinski problem). See Alex's answer for more examples.
Even more reading The Sierpiński Problem: Definition and Status.
A: Carrying forward the observations of @lulu: 
If $n = 0$ (mod 8), then $19| a_n$ for $n=8, 80, 152, 224,...$ i.e. for $8$ and every ninth multiple of $8$. 
For since $2^8=9$ (mod $19$), and $3061=2$ (mod $19$), then $2^8\cdot3061+1=9\cdot2+1=0$ (mod 19). Further, since $2^{72}=1$ (mod $19$), then $19|a_n$ for $n=8, 80, 152, 224,...$.
In the same way we find that $31|a_n$ for $n=32, 72, 112, 152,...$, i.e. for every fifth multiple of $8$ beginning with $32$.   
$37|a_n$ for $n=48, 120, 192, 264,...$, or every ninth multiple of $8$ from $48$.
$43|a_n$ for $n=32, 88, 144, 200,...$,i.e. for every seventh multiple of $8$ beginning with $32$.
Once more, $53|a_n$ for $n=80, 184, 288, 392...$, or every thirteenth multiple of $8$ from $80$.
Thus, even allowing for some overlap, it seems that with just these five primes we sieve out$$\frac19+\frac15+\frac19+\frac17+\frac{1}{13}>\frac12$$of $a_n$ as necessarily composite.
If we did not have an accepted answer to the question, this could have at least further narrowed the search.      
