TITLE: Find the least positive integers $n>1$ such that $n\mid 2^{n-1}+3$
QUESTION [6 upvotes]: Find the smallest integers $n>1$ such that $n\mid 2^{n-1}+3.$

I find $n=13957196317$ works, but I wonder if this is the smallest one.
Here is my method:
If $n$ is prime, then $2(2^{n-1}+3)\equiv 2^n+6\equiv 2+6\equiv8 \mod n,$ hence $n=2,$ false.
If $n=pq$ where $p,q$ are both odd prime numbers, then
$$2^{n-1}+3\equiv 2^{p-1}+3\equiv 0 \mod q\\2^{n-1}+3\equiv 2^{q-1}+3\equiv 0 \mod p,$$
factorizing $2^{p-1}+3$ we get some prime factors $q$, if $p\mid 2^{q-1}+3$ then $n=pq$ is a solution.
In this way I get $n=61\times 228806497$ and $n=67\times 44210291368986343.$
PS: We can prove that $GCD(n,30)=1$. 
There is no solution up to $n=10^8$, this is my Mathematica code(up to $n=10^7$):
Do[If[GCD[n,15]==1 && PowerMod[2,n-1,n]==n-3, Print[n]; Break[]], {n,3,10^7,2}]


Updates:
Gammatester says no solutions till $n\lt2^{31}-1$

REPLY [2 votes]: This answer on my SO post, has confirmed that $13957196317$ is indeed the first value.