Firstly, we prove that if $m$ satisfies the first condition, then $m$ has only one prime divisor. Assume that $m$ has at least 2 prime divisors, let $p$ be the smallest prime divisor of $m$ and $p_1, p_2, \dots, p_k$ be the remaining prime divisors of $m$. We have
$$\frac{\varphi(m)}{m} = \left(1 - \frac{1}{p}\right) \left(1 - \frac{1}{p_1}\right) \dots \left(1 - \frac{1}{p_k}\right) < 1 - \frac{1}{p} = \frac{p-1}{p}.$$
Hence, by choosing $n = p$ in i), one can get
$$\frac{s(p, m)}{m-p} = \frac{\varphi(m) - (p-1)}{m-p} < \frac{\varphi(m) - \frac{\varphi(m)}{m} \cdot p}{m-p} = \frac{\varphi(m)}{m} = \frac{s(1, m)}{m},$$
which is a contradiction. Therefore, $m$ must be a power of a prime. Let $m = p^k$, note that
$$2022^{p^k} + 1 \equiv 2022 + 1 \equiv 2023 \equiv 0 \pmod{p},$$
thus $p \mid 2023$ and $p \in \{7, 17\}$.
If $p = 7$, using LTE, we have
$$v_7(2022^{7k} + 1) = v_7(2023) + v_7(7^k) = 1 + k \ge v_7(7^{2k}) = 2k.$$
From this, we conclude that $k = 1$ and $m = 7$. Similarly, for $p = 17$, applying LTE, we also have
$$v_{17}(2022^{17k} + 1) = v_{17}(2023) + v_{17}(17^k) = 2 + k \ge v_{17}(17^{2k}) = 2k,$$
so $k \in \{1, 2\}$ and $m \in \{17, 289\}$.
Therefore, $m = 7, 17, 289$ are all desired numbers. $\square$