You're trying to PROVE that $G$ is cyclic, so you cannot (yet) assert that $x = g^n$. Instead, consider $x \cdot x$. It must be either $x, y,$ or $e$. If it's $x$, then you have
$$
x^2 = x\\
x^2 (x^{-1}) = x x^{-1}\\
x = e
$$
which is a contradiction, because $x$ and $e$ are distinct elements of the group.
If $x^2 = e$, then $x$ has order 2, but 2 does not divide 3, so this contradicts Lagrange's theorem.
Finally, we conclude that $x^2 = y$, and thus the group is cyclic, generated by the element $g = x$.
{Alternative if you don't like Lagrange yet:}
In the case where we suppose that $x^2 = e$:
The elements $xe, xx,$ and $xy$ must all be distinct for if two were the same, then multiplying by $x^{-1}$ on the left would show that two of $e, x, y$ were the same, which is impossible.
Since $xe = x$ and we're assuming $x^2 = e$, we must have
$$
xy = y.
$$
multiplying on the right by $y^{-1}$ gives $x = e$, a contradiction. So $x^2 = e$ is also impossible.
Suppose $G$ is a cyclic group of order $n$, then there is at least one $g \in G$ such that the order of $g$ equals $n$, that is: $g^n = e$ and $g^k \neq e$ for $0 \leq k < n$. Let us prove that the elements of the following set
$$\{g^s \: \: \vert \: \: 0 \leq s < n, \text{gcd}(s,n) = 1\}$$
are all generators of $G$.
In order to prove this claim, we need to show that the order of $g^s$ is exactly $n$. Suppose that it is $k$, where $0 < k \leq n$. We have that
\begin{equation}
(g^s)^n = (g^n)^s = e
\end{equation}
and therefore we must have that $k$ divides $n$. Let us now prove that $n$ divides $k$. Because of Euclid's lemma, there are $q, r \in \mathbb{N}$ such that $k = qn + r$, where $0 \leq r < n$. We have that
\begin{equation}
e = (g^s)^k = (g^s)^{qn} \cdot (g^s)^r = (g^s)^r = g^{sr}.
\end{equation}
Because the order of $g$ is $n$, we must have that $n$ divides $sr$. However, because $\text{gcd}(s,n) = 1$, we must have that $n$ divides $r$, but this would mean that either $n \leq r$ (impossible because of $0 \leq r < n$) or $r = 0$. Since $r = 0$ is the only possibility, we have that $k = qn$, so $n$ divides $k$ and therefore we must have that $k = n$. So $g^s$ is a generator of $G$ in the case that $\text{gcd}(s,n) = 1$.
This proves the claim made in the answer of E.Joseph, that there are exactly $\varphi(n)$ generators (since $\varphi(n)$ is exactly the number of elements which are coprime to $n$). It also gives you an idea on how to find all generators, given that you know one generator.
Best Answer
For the trivial group, and the group with two elements, there is one generator. So the answer to your question is yes if $n=1$.
For cyclic groups of order larger than $2$, if $x$ is a generator, then $x^{-1}$ is also a generator, and $x\ne x^{-1}$ else $x$ has order $1$ or $2$ (contradiction). Hence, in this case, generators come in pairs and hence there must be an even number of them. So the answer to your question is no if $n>1$ and $n$ is odd.
If $n$ is even, then as the comments note, a cyclic group of order $m$ has $\varphi(m)$ generators, so we need to find some $m$ such that $\varphi(m)=n$. However, it is not always the case that such $m$ exists. Even $n$ that are not equal to $\varphi(m)$ for some $m$ are called "non-totients" and are listed in this sequence. So finally, the answer to your question is no if $n$ is even and a non-totient, and yes if $n=\varphi(m)$ for some $m$.