For which values of $k$ is it known that there are infinitely many $n$, such that $2^{n+k}\equiv 1\pmod{n}$?

For any $k\geq 1$, there are infinitely many solutions of the congruence $2^{n+k}\equiv 1\pmod{n}$. To see this, observe first that there is always a solution $n\geq 1$ satisfying $n+k\geq 7$. Indeed, for $k\geq 6$ this is verified by the trivial solution $n=1$, while for $1\leq k\leq 5$ it is verified by the pairs $$ (k,n) \ = \ (1,15),\ \ (2,7),\ \ (3,5),\ \ (4,31),\ \ (5,3).$$ Now it suffices to show that, for any fixed $k$ and for any solution $n\geq 1$ satisfying $n+k\geq 7$, there is a bigger solution $n'>n$ for the same $k$. Indeed, let $p$ be a primitive prime divisor of $2^{n+k}-1$, which exists by Zsigmondy's theorem. Then, the order of $2$ modulo $p$ equals $n+k$, whence $n+k\mid p-1$. This implies that $pn+k=(p-1)n+(n+k)$ is divisible by $n+k$, hence $$p,n\mid 2^{n+k}-1\mid 2^{pn+k}-1.$$ However, $p>n+k>n$, so $p$ and $n$ are coprime, and the above implies that $pn\mid 2^{pn+k}-1$. That is, $n'=pn$ is a solution bigger than $n$.

Sorry, this is in fact trivial. For instance if $2^{n+1}\equiv1\pmod{n}$ then $n$ is of course odd, so $2^{3n+3}\equiv1\pmod{n}$ and mod $3$, so $2^{3n+3}\equiv1\pmod{3n}$. This probably works for all odd $k$ not only $3$.