Starting with the case where you have $17 and 8 flips left trying to get to 21 knowing the coin is fair (betting below 1/2 on the first 2 flips and playing on from there with a coin of potentially indeterminate probability is a pain in the ass):

The "efficient" amounts of money you can have with n flips left are 0/21..2^n/21. With no flips left, 0 or 21. With 1 flip left, 0, 10.5, or 21. With 2 flips left, 0, 5.25, 10.5, 15.75, 21, etc. and the strategies from these amounts are obvious. Any extra money in between the efficient amounts is completely useless and can be ignored, so your odds of making 21 with X money and N flips to go are int((2^n)*(X/21))/2^n (by knowing you end at 0 or 21 and that the flips are neutral EV).

So that comes out to 207/256, as does 4 8 5 10 1 2 4 8.

I would say there are some other answers that are equally correct and could have some advantages, like:
Bet all on same side every roll

It is simpler solution - actually, as simple as it can be: pick whichever side you want (head or tail), then bet all on that side in every roll from 1st to 10th. Beside being simpler, it also result in higher payout amount after win ( although with same EV, so wins would be twice less often )

And it is fairly easy to show that it reduces to same solution as yours. Assume EV=evA in your "bet 0 on first roll, then bet all in rolls 2-10 on whatever dropped on first roll (heads or tails)", then:
1) if we expand your solution to bet all on first roll (on any side, head or tails, select randomly), we will win in 50% (increasing our starting amount to $40), or lose in 50% (and stop our rolls). So, if we won first roll and continue to implement your solution, our EVwon = 2*evA (due to $40), but since there is only 50% chance we will win, our actual EV= 50%*2*evA == evA ... in other words, exactly same as for your original approach
2) since we will always continue to bet on same coin side (head or tails) in our 2nd-10th roll as in that 1st roll, it practically means if we randomly pick side and bet on it in all 10 rolls, we get same EV as in your approach

Actual EV, assuming p=probability for biased coin to drop on one side , is :
EV= $20*2^9*(p^10+(1-p)^10)

That part with (1-p)^10==q^10 is for rare cases when we select wrong side and manage to roll it 10 times anyway. If any coin bias is equally probable (unnecessary assumption, but just for giving example), than integral(p=0..1)(p^10+q^10)~0.18, and EV~$1850. Anyway, EV is identical for your or mine solution, although it can be expressed differently:
a) in your solution, EV= $10240 with chance (p^10+q^10) ( or ~18% on average)
b) in my solution, EV = $20480 with chance (p^10+q^10)/2 ( or ~9% on average)

Since any solution with same EV is not clearly 'better' and will depend on subjective risk vs reward preferences, my point was not that my solution is better than yours - although higher payout may be important in some cases, and there are certainly solutions on other side of spectrum, with lower payout but higher chances than yours. My point was rather to object to your "I don't think any other answer is correct" statement. If you have used "better" instead of "correct", that could have been acceptable under 'subjective risk vs reward opinion' ... but "correct" has fairly precise meaning, and any solution resulting in same 'best' EV should be considered 'correct'.

Actual chance of 1 2 4 8 5 10 1 2 4 8 to win $21, with specified biased coin probability p (p>=q), and assuming betting always on side that dropped most often, is:
p21(p)=1-(1-p)^2/8*(2-p+p*(1-p)^2/4)

For fair coin, p21(0.5)= 0.9521484375
For very biased coin, p21(0.9)= 0.9986221875
For totally biased coin, p21(1)= 1 (guaranteed $21 in two rolls)

Your int((2^n)*(X/21))/2^n formula for X=$20 and n=10 gives odds to get $21 as 975/2^10 = 0.9521484375 , and since it is identical to p21(fair coin) for 1 2 4 8 5 10 1 2 4 8 , it follows that 1 2 4 8 5 10 1 2 4 8 is optimal solution for fair coin.

But it is still open question if it is also optimal for biased coin, although I was not able to find any other approach that result in better odds. For example, 0000001248 approach (make sure if heads or tails is favored, then use it) has 'only' 0.9895024 chance to win $21 for biased p=0.9, which is less than 0.9986221875 with 1 2 4 8 5 10 1 2 4 8 - so obviously starting as soon as possible to bet with chance for $21 is better approach even with biased coin.