Solution:
Let $m = n - 9$ and for the sake of contradiction assume $m$ has a prime factor $p$ with $p \neq 3$. Express $p = 9q + r$ where $0 \leq r \leq 8$ ($r$ is the remainder when $p$ is divided by $9$ and $q$ may be zero). Since $\gcd(9, p) = 1$ we must have $r = 1, 2, 4, 5, 7$ or $8$. For each of these possibilities we will use a different choice for $x$ to get our contradiction.

if $r = 1$ then choose $x = q$ so that $(9x + 1) = 9q + 1 = p$

if $r = 2$ then choose $x = 5q + 1$ so that $(9x + 1) = 45q + 10 = 5p$

if $r = 4$ then choose $x = 7q + 3$ so that $(9x + 1) = 63q + 28 = 7p$

if $r = 5$ then choose $x = 2q + 1$ so that $(9x + 1) = 18q + 10 = 2p$

if $r = 7$ then choose $x = 4q + 3$ so that $(9x + 1) = 36q + 28 = 4p$

if $r = 8$ then choose $x = 8q + 7$ so that $(9x + 1) = 72q + 64 = 8p$

In any we can choose $x$ so that $(9x + 1)$ is a multiple of $p$. For this particular value of $x$ we have
$$(n x + 1) = m x + (9 x + 1),$$
and so $(n x + 1)$ is a multiple of $p$ too (recall $p \mid m$). Therefore $p$ would be a common factor of $(9x + 1)$ and $(n x + 1)$. This is a contradiction so no such $p$ can exist.

So $m = n - 9$ cannot have any prime factors other than $3$. Hence $m = 3^{k}$ or $m = -3^{k}$ for some integer $k \geq 0$. i.e.
$$n = 9 + m = 9 + 3^{k} \text{ or } 9 - 3^{k}$$
The positive integers of this form, less than $2024$ are: $6, 8, 10, 12, 18, 36, 90, 252, 738$.

To show all these work, we now assume $n = 9 \pm 3^{k}$ for some integer $k \geq 0$, and let $g = \gcd(9x + 1, nx + 1)$. Since $9x + 1$ is not a multiple of $3$, we cannot have $g$ being a multiple of $3$. However
$$g \mid (n x + 1) - (9 x + 1) = \pm 3^{k}.$$
The only divisors of $3^{k}$ which are not a multiple of $3$ are $1$ and $-1$. Therefore $g = 1$ whenever $n = 9 \pm 3^{k}$.