Solution:

For a nice number $n$ and a given partition of the set $\{1,2, \ldots, n\}$ into two-element subsets such that the sum of the elements in each subset is a power of $3$, we say that $a, b \in \{1,2, \ldots, n\}$ are paired if both of them belong to the same subset.

Let $x$ be a nice number and $k$ be a (unique) non-negative integer such that $3^{k} \leq x < 3^{k+1}$. Suppose that $x$ is paired with $y < x$. Then, $x + y = 3^{s}$, for some positive integer $s$. Since
$$3^{s} = x + y < 2x < 2 \cdot 3^{k+1} < 3^{k+2}$$
we must have $s < k+2$. On the other hand, the inequality
$$x + y \geq 3^{k} + 1 > 3^{k}$$
implies that $s > k$. From these we conclude that $s$ must be equal to $k+1$, so $x + y = 3^{k+1}$. The last equation, combined with $x > y$, implies that $x > \frac{3^{k+1}}{2}$.

Similarly as above, we can conclude that each number $z$ from the closed interval $[3^{k+1} - x, x]$ is paired with $3^{k+1} - z$. Namely, for any such $z$, the larger of the numbers $z$ and $3^{k+1} - z$ is greater than $\frac{3^{k+1}}{2}$ which is greater than $3^{k}$, so the numbers $z$ and $3^{k+1} - z$ must necessarily be in the same subset. In other words, each number from the interval $[3^{k+1} - x, x]$ is paired with another number from this interval. Note that this implies that all numbers smaller than $3^{k+1} - x$ are paired among themselves, so the number $3^{k+1} - x - 1$ is either nice or equals zero. Also, the number $3^{k}$ must be paired with $2 \cdot 3^{k}$, so $x \geq 2 \cdot 3^{k}$.

Finally, we prove by induction that $a_{n} = 2^{n} - 1$, where $a_{n}$ is the number of nice positive integers smaller than $3^{n}$. For $n = 1$, the claim is obviously true, because $2$ is the only nice positive integer smaller than $3$. Now, assume that $a_{n} = 2^{n} - 1$ for some positive integer $n$. We will prove that $a_{n+1} = 2^{n+1} - 1$. To prove this, first observe that the number of nice positive integers between $2 \cdot 3^{n}$ and $3^{n+1}$ is exactly $a_{n+1} - a_{n}$. Next, observe that $3^{n+1} - 1$ is nice. For every nice number $2 \cdot 3^{n} \leq x < 3^{n+1} - 1$, the number $3^{n+1} - x - 1$ is also nice and is strictly smaller than $3^{n}$. Also, for every positive integer $y < 3^{n}$, obviously there is a unique number $x$ such that $2 \cdot 3^{n} \leq x < 3^{n+1} - 1$ and $3^{n+1} - x - 1 = y$. Thus,
$$a_{n+1} - a_{n} = a_{n} + 1 \Leftrightarrow a_{n+1} = 2 a_{n} + 1 = 2(2^{n} - 1) + 1 = 2^{n+1} - 1$$
completing the proof.

In summary, there are $2^{2022} - 1$ nice positive integers smaller than $3^{2022}$.