<s>[INST] Can I find all solutions of $2^{n-1}\equiv k\mod n$? Suppose$\ k\ge 2\ $ is a positive integer.

Can I find all positive integers $\ n>1\ $ with $$2^{n-1}\equiv k\mod n$$ ?

I only found out yet that there is always a solution if $\ k>2\ $ and $\ k-1\ $ is not a power of $\ 2\ $. In this case, $\ k\ $ has an odd prime factor $\ q\ $, for which we have $\ 2^{q-1}\equiv k\mod q\ $ as desired.
I am particularly interested whether for $\ k=5\ $, there is a solution and whether for $\ k=11\ $, there is a solution besides $\ n=5\ $. Finally, for $\ k=3\ $, is $\ 10669\ $ the only solution?
 [/INST] Clearly n can not be prime. I had following experiment:
$2^{4-1}=8=2\times 4 +0$
$2^{6-1}=32=5\times 6 +2$
$2^{8-1}=128=16\times 8+0$
$2^{9-1}=256=28\times 9 +4$
$2^{10-1}=512=51\times 10+2$
$2^{12-1}=2042=170\times 12 +8$
$2^{14-1}=8192=585\times 14 +2$
$2^{15-1}=16384=1092\times 15 +4$
$2^{16-1}=32768=2048\times 16 +0$
$2^{17-1}=65536=3855\times 17 +1$
$2^{18-1}=131072=7281\times 18+14$
$2^{33-1} ≡4 \mod 33$
$2^{27-1} ≡13 \mod 27$ </s>