You've made a good start, but note that, as explained below, the factors of $2022 - 1 = 43(47)$ don't apply. To finish, first define
$$\alpha(n,m)=|\{1\le k\le n-1 : (k,m)=1\}| \tag{1}\label{eq1A}$$
This then gives
$$\begin{equation}\begin{aligned}
\phi(n,m) & = |\{1\le k\le m : (k,m)=1\}| - |\{1\le k\le n-1 : (k,m)=1\}| \\
& = \phi(1,m) - \alpha(n,m)
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Since $\phi(1,m)=\phi(m)$, as you've noted, then using \eqref{eq2A}, the first problem condition becomes
$$\begin{equation}\begin{aligned}
\frac{\phi(m) - \alpha(n,m)}{m-n} & \ge \frac{\phi(m)}{m} \\
m(\phi(m) - \alpha(n,m)) & \ge (m - n)(\phi(m)) \\
-m\alpha(n,m) & \ge -n\phi(m) \\
\frac{\alpha(n,m)}{n} & \le \frac{\phi(m)}{m}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Let $p_1$ be the smallest prime factor of $m$. If $m$ has at least $2$ distinct prime factors then, with $n = p_1$, the left side of \eqref{eq3A} becomes $\frac{p_1-1}{p_1} = 1 - \frac{1}{p_1}$. However, using Euler's product formula, the right side of \eqref{eq3A} becomes $\prod_{p\mid m}\left(1 - \frac{1}{p}\right) \lt 1 - \frac{1}{p_1}$, contradicting the inequality in \eqref{eq3A}.
Thus, $m$ can have only one distinct prime factor, so for some integer $a \ge 1$,
$$m = p_1^a \tag{4}\label{eq4A}$$
The RHS of \eqref{eq3A} is thus $1 - \frac{1}{p_1}$. For the LHS, consider $n \gt 0$ with $n = sp_1 + t$, $0 \le s \lt p_1^{a-1}$ and $0 \le t \lt p_1$. With $t = 0$, then $s \ge 1$ with there being $s-1$ multiples of $p_1$ less than $n$, which means $\frac{\alpha(n,m)}{n} = \frac{(n - 1) - (s - 1)}{n} = \frac{n-s}{n} = 1 - \frac{s}{sp_1} = 1 - \frac{1}{p_1}$, so \eqref{eq3A} holds. Alternatively, with $1 \le t \le p_1 - 1$, then $\frac{\alpha(n,m)}{n} = \frac{(n - 1) - s}{n} = 1 - \frac{s + 1}{sp_1 + t} \lt 1 - \frac{s + 1}{sp_1 + p_1} = 1 - \frac{s+1}{(s+1)p_1} = 1 - \frac{1}{p_1}$, so \eqref{eq3A} also holds in this case.
This means all $a \ge 1$ satisfy \eqref{eq4A}. Next, the second required condition is
$$m^2 \mid 2022^m + 1 \; \to \; m^2 \mid 2022^{2m}-1 \tag{5}\label{eq5A}$$
Thus, the multiplicative order of $2022$ modulo $p_1$ must divide $2m$, i.e.,
$$(r = \operatorname{ord}_{p_1}(2022)) \mid 2p_1^{a} \tag{6}\label{eq6A}$$
However, as you've already indicated, since $2022^{p_1-1}\equiv 1\pmod{p_1}$, then $r \mid p_1 - 1$. Thus, as you've already stated, from \eqref{eq6A} we have $r \mid (\gcd(2p_1^{a},p-1) = 2)$, i.e., $r$ must be $1$ or $2$. It can't be $1$ since then $p_1 \mid 2022^{m} - 1$, contradicting the LHS of \eqref{eq5A}, so it must be $2$. This means
$$p_1 \mid (2022 - 1)(2022 + 1) \tag{7}\label{eq7A}$$
We can't have $p_1 \mid 2022 - 1$ since then $r = 1$, which has already been discarded, so $p_1 \mid 2022 + 1$. Next, the Lifting-the-exponent lemma, \eqref{eq4A} and the LHS of \eqref{eq5A} gives
$$\begin{equation}\begin{aligned}
\nu_{p_1}(2022^{m}+1) & = \nu_{p_1}(2022 + 1) + \nu_{p_1}(m) \\
& = \nu_{p_1}(2022 + 1) + a \\
& \ge 2a
\end{aligned}\end{equation}\tag{8}\label{eq8A}$$
This means $\nu_{p_1}(2022 + 1) \ge a$. Since $2022 + 1 = 7(17^2)$, we have either $p_1 = 7$, which gives $m = 7$, or $p_1 = 17$, which gives $m = 17$ or $m = 17^2 = 289$. Thus,
$$m \in \{7, 17, 289\} \tag{9}\label{eq9A}$$
are the only values of $m \ge 2$ which satisfy both of the problem conditions.
Best Answer
If a sequence of consecutive integers sums to a given number, and if the number of members of that sequence is odd, the average of the first and last members will be a whole number which is equal to the median number of the sequence. If the number of members is even, then the average of the first and last numbers will be a half integer disposed between the two central numbers in the sequence. Moreover, if the number of members of the sequence is odd, then the number of members must be a factor of the sum. If the number of members of the sequence is even, then the number of members is twice a factor of the sum. Finally, if the sum is to include only positive integers (NB the question stipulates integers, without limitation), the number of members of the sequence must be smaller than the square root of the sum.
$2022=2\cdot3\cdot337$. $2022$ has $8$ factors: $1,2,3,6,337,674,1011,2022$
The factor $1$ (formally, at least) gives rise to a sequence with one member: $2022$
The factor $2$ gives direct rise to no sequences, because $\frac{2022}{2}=1011$ and there can be no sequence with an even number of members centered on a whole number. However, $2$ gives indirect rise to a sequence because $\frac{2022}{2\cdot 2}=505.5$ which is a half integer, so we can construct a sequence of four members surrounding $505.5$, viz: $504,505,506,507$
The factor $3$ gives direct rise to a sequence with $3$ members centered on $\frac{2022}{3}=674$, viz: $673,674,675$.
The factor $6$ gives direct rise to no sequence because $\frac{2022}{6}=337$, a whole number which cannot be the center of an even number of members. But it gives indirect rise to a sequence of $12$ members because $\frac{2022}{12}=168.5$, viz: $163,164,165,166,167,168,169,170,171,172,173,174$
The four sequences described to this point (except the single member sequence $2022$) have all been identified by previous responders. Next, we examine instances in which the number of members of the sequence is greater than the square root of the sum.
The factor $337$ gives direct rise to a sequence because $\frac{2022}{337}=6$, so there is a sequence of integers with $337$ members centered on $6$ that sums to $2022$. However, many of the members of that sequence would be negative integers or zero. I will omit a full listing of that sequence here, but it begins at $-162$ and runs to $174$. Every positive integer smaller than $163$ is cancelled with a corresponding negative integer, leaving only $163,\dots,174$ already identified above.
Also, there will be a sequence of $1348$ members centered on $1.5$ and a sequence of $4044$ members centered on $0.5$. That final sequence is $-2021,-2020,-2012,\dots,-1,0,1,2,\dots,2021,2022$. Here, every positive integer except $2022$ is matched with a corresponding negative integer, cancelling each other out and leaving the sum as $2022$
Since the answer that OP stated as correct was $4$, I presume that the question intended to limit the sequences to positive integers.