If $x^m=e$ has at most $m$ solutions for any $m\in \mathbb{N}$, then $G$ is cyclic

The proof that I know of this fact goes as follows. Say that $|G|=n$ and for $1\leq d\leq n$ let $$ A_d=\{g\in G: g \text{ has order } d\}. $$ Then one proves:

(1) that $A_d=\emptyset$ if $d$ is not a divisor of $n$;

(2) that if $d\mid n$, then $|A_d|\leq\varphi(d)$.

This allows us to conclude because the chain of inequalities $n=|G|=\sum_{d\mid n}|A_d|\leq\sum_{d\mid n}\varphi(d)=n$ gives actual equality at every step and, in particular, $A_n\neq\emptyset$, i.e. $G$ admits a generator.

There are many ways to proceed with this problem, so I give one which I don't remember seeing in a text, though perhaps it uses a little more knowledge of group theory. We proceed by induction. The result is true if $|G| =p$ for some prime $p$, (and if $|G| =1$), so suppose that $|G| >1$ is not prime and the result is true for groups of order less than $|G|$. If $H$ is any proper subgroup of $G$, then $H$ is cyclic. If $H$ has order $m$, then $H$ contains $m$ solutions in $G$ of $x^m = 1$. But there are only m solutions of $x^m = 1$ in $G$, so $H = \{ x \in G: x^m = 1 \}$. Notice that if $x^m = 1$ then $(gxg^{-1})^m = 1$ for any $g \in G$. Hence $gHg^{-1} = H$ for any $g \in G$. Hence every subgroup of $G$ is normal. In particular, each Sylow $p$-subgroup of $G$ is normal in $G$, and $G$ is the direct product of its Sylow $p$-subgroups. If all Sylow $p$-subgroups of $G$ are proper, then all are cyclic by the induction hypothesis, and then $G$ itself is cyclic. Hence we may suppose that $G$ is a $p$-group for some prime $p$. Let $p^e$ be the largest order of an element of $G$. Then $G$ has a cyclic subgroup, say $N$, of order $p^e$, and $N \lhd G$ as above. But choose $x \in G$. Then $x$ has order $p^f$ for some non-negative integer $f$. If $f >e$, the choice of $e$ is contradicted. If $f \leq e$, then $x \in N$, since $x^{p^e} = 1$ and $N$ contains all solutions in $G$ of $y^{p^e} = 1$. Hence $N = G$ and $G$ is cyclic.

Please refer I.N. Herstein " Topics in Algebra" Second Edition. Page $358$. Chapter: Selected Topics.

He solves it by considering different cases. The solution given in his book is as follows:

If the order of $G$ is a power of some prime number $q$, then the result is very easy. For suppose that $a \in G$ is an element whose order is as large as possible: its order must be $q^{r}$ for some integer $r$. The elements $e,a, a^{2},\cdots, a^{q^r-1}$ give us $q^{r}$ distinct solutions of the equation $x^{q^r}=e$, which by our hypothesis implies that these are all the solutions of this equation. Now if $b \in G$, its order is $q^{r}$ where $s \leq r$, hence $$b^{q^r} = (b^{q^s})^{q^{r-s}}=e$$ By the observation made above this forces $b=a^{i}$ for some $i$ and so $G$ is cyclic.

The general finite abelian group $G$ can be realized as $G=S_{q_1} \cdot S_{q_2} \cdots S_{q_k}$ where the $q_i$ are the distinct prime divisors of $|G|$ and where the $S_{q_i}$ are the sylow subgroups of $G$. Moreover, every element of $g \in G$ can be written in a unique way as $g=s_{1}s_{2}\cdots s_{k}$ where $s_{i} \in S_{q_i}$. An solution of $x^{n}=e$ in $S_{q_i}$ is one of $x^{n}=e$ in $G$ so that each $S_{q_i}$ inherits the hypothesis we have imposed on $G$. By the remarks of the first paragraph of the proof, each $S_{q_i}$ is a cyclic group; let $a_i$ be a generator of $S_{q_i}$. We claim that $$c=a_{1}\cdot a_{2} \cdots a_{k}$$ is a cyclic generator of $G$. To verify this, all we must do is to prove, that $|G|$ divides $m$, the order of $c$. Since $c^{m}=e$, we have that $a_{1}^{m}\cdot a_{2}^{m} \cdots a_{k}^{m}=e$. By the uniqueness of representation of an element of $G$, as a product of elements in the $S_{q_i}$ we conclude that each $a_{1}^{m}=e$. Thus $|S_{q_i}| \mid m$ for every $i$. Thus $$|G| = |S_{q_1}| \cdot |S_{q_2}| \cdots |S_{q_k}| \ \Bigl|\: m$$ However $m \mid |G|$ and so $|G|=m$. This proves that $G$ is cyclic.