It is not possible to do it in 1 question, but can easily be done in 2.
@hexomino proposes a solution with 1 question, but it does not work because:
The question is not well defined.
If we ask a knave to point at the door with the pot of gold, he has more than one option. So if we ask a knight or knave what the knave would have done, he can't know for sure, so he can't answer the question. Even if we somehow assume that they know which wrong answer the knave would have given, the knave still has the choice to give the other wrong option.
More generally, we cannot do it in 1 question because:
Every question can only give us (at best) binary information, unless we know we are talking to a knight.
If we ask a question with more than 2 possible answers, a knave could pick 2 options in advance, and answer the first wrong one. So if you are talking to a knave, you can get binary information at best. Since you cannot know whether you are talking to a knave, you can't do better than that.
A more general solution to the problem of arbitrarily many knights and knaves (even without knowing the number of knights and knaves) is the following one:
It is always possible in log2(number of doors) questions.
The questions can be of the form:
"If the roles of knights and knaves were inverted, what would you tell me if I asked..."
With a simply binary decomposition, we can then get the information we want.
The only situation with a faster solution is:
If it is faster to find a knight than to find the answer, we can find a knight, then ask him the question directly. For example if there are 10 knights and 1 knave, with 11 doors, 1 question is enough to find a group with the knave, and a second question to find the door, where as the normal approach would need 4 questions.
Edit: after further reflections and a comment from @hexomino, I came up with a 1 question solution that might work (if we admit it):
"Please point to a door B could have pointed to, if I had asked him to point at a door C could have pointed to if I had asked him to point at the door with the pot of gold."
This works assuming knaves have to point at a door, and we count this as a question (it's more of an order than a question).
My proof fails, because here there is more than 1 correct answer at some point, so a knave cannot pick 2 answers, and answer the first wrong one among the 2.
Open questions like this make it unclear what "lying" means (for example a knave could point at himself instead of pointing at a door). I would say this is the reason the problem is generally formulated as asking yes/no questions, with knights giving the correct answer, and knaves the other one, makes the problem a lot better defined.