Let's observe following matrix with an infinite number of elements :
Elements of the main diagonal are of the form $n\cdot2^n+1$ . These numbers are known as Cullen numbers . It is an open question whether there is an infinite number of Cullen primes. In fact Cullen prime is special case of prime number of the $k\cdot 2^n +1$ form.
So, maybe easier question is : Are there infinitely many primes of the form $k\cdot 2^n +1$ ?
Best Answer
The first row of the OP's matrix is 3,5,7,..., and so obviously contains every prime except 2. More generally, Dirichlet's theorem implies that each row contains an infinite number of primes (as pointed out by marty cohen).
A Sierpinski number is a odd positive integer $k$ such that $k \cdot 2^n +1$ is composite for all $n \geq 1$. The smallest known Sierpinski number is $78557$ [the Seventeen or Bust team are trying to (computationally) prove it is the smallest]. Sierpinski showed that there are an infinite number of Sierpinski numbers, so there are an infinite number of all-composite columns in the OP's matrix (which extends zyx's answer).
Numbers of the form $k \cdot 2^n+1$ with $k<2^n$ are (sometimes) referred to as Proth numbers. Proth's Theorem gives a necessary and sufficient condition for the primality of a Proth number. Yves Gallot proth.exe is an available implementation of Proth's Theorem. Many of the largest known primes are Proth primes.
Related are numbers of the form $k \cdot 2^n-1$, for which we can detect primality via the efficient Lucas-Lehmer-Riesel test, implemented by Jean Penné's LLR. Also, there's Mersenne numbers, for which there's the giant GIMPS project searching for large Mersenne primes.