Computerless proof that CDJB's solution is best possible (which also finds all possible solutions in that many steps, though I haven't filled in the details of that part since you can see the explicit solutions in CDJB's answer):
We can get to 9 only from 27, 28, or 29. (One step.) We can get to those from anything in [81,89] -- this notation denotes a closed interval, including its endpoints -- or by doubling 14. (Two steps.) We can get to those from anything in [243,269], or anything in [41,44], or by doubling 7. (Three steps.) We can already see a solution by doubling 8 times to reach 256. We don't have enough doublings available to get into [243,269] in fewer than 8 more steps, so if there's a solution that uses fewer than 11 in total it must go via [41,44] or 7. We can get to those from anything in [123,134] or [21,23] or by doubling 21 or 22 -- note that those last two don't add anything to the [21,23] we already had. (Four steps.) Again we can see that we could go via 128 in 11 steps; if we use fewer steps we can't reach [123,134] with the number of doublings available, so we'd have to go via [21,23]. We can get to those from anything in [63,71] or by doubling 11. (Five steps.) Again we could use 64, and now everything that's left is out of range if we use <=5 more steps, since that can't take us above 32. Done.
Credit where due: isaacg pointed out, in comments, a couple of bugs in my analysis, and while fixing those I found another one. Thanks, isaacg!