Solution:

Let $k$ be the number of golden terms. We claim that $k \leq 1010$.

Proof: Define the term immediately before a golden term to be a silver term. The number of silver terms is also $k$. If $a$ is any silver term and $b$ is the corresponding golden term then we must have
$$a \leq \frac{b}{2} \leq \frac{2021}{2} < 1011.$$
Therefore $a \leq 1010$. Hence $a \in \{1,2,3,\ldots ,1010\}$ and therefore there can be at most 1010 different silver terms. Therefore $k \leq 1010$.

Now we need to show that $k = 1010$ is indeed possible. To do this we first partition the set $\{1,2,3,\ldots ,2021\}$ into the following distinct parts:
$$\{1,2,4,8,16,32,64,128,256,512,1024\},$$
$$\{3,6,12,24,48,96,192,384,768,1536\},$$
$$\{5,10,20,40,80,160,320,640,1280\},$$
$$\{7,14,28,56,112,224,448,896,1792\},$$
$$\ldots$$
$$\{2019\},$$
$$\{2021\}.$$
Each part starts with an odd number, which is then doubled until the result is larger than 2021.

Putting each part in the above sequence in order is our example. In this example every even term, $2n$, is golden because it occurs immediately after the term $n$. There are 1010 even numbers, so $k = 1010$ is possible.