How to prove this conjecture $x_{n}(n\ge 4)$ can't be an integer. 
Define sequence $x_{1}=1$,and $x_{n+1}=1+\dfrac{n}{x_{n}},n\ge 1$.
  I found
  $$x_{2}=2,x_{3}=2,x_{4}=1+\dfrac{3}{x_{3}}=\dfrac{5}{2}, x_{5}=1+\dfrac{4}{2.5}=\dfrac{13}{5},x_{6}=1+\dfrac{5}{x_{5}}=\dfrac{38}{13}$$
  $$x_{7}=1+\dfrac{6}{x_{6}}=\dfrac{58}{19},x_{8}=1+\dfrac{7}{x_{7}}=\dfrac{191}{58},\cdots$$

I conjecture:   $x_{n}\notin Z$ for $n\ge 4$; in other words, this sequence has only three integer terms. 
 A: The conjecture is true.
Define, as in @S.C.B.'s comment, the sequence of Telephone numbers:
$$
T_0 = T_1 = 1, \quad T_{n+1} = T_{n} + nT_{n-1},\ n\ge 1, \tag{1}
$$
so that 
$$
x_n = \frac{T_n}{T_{n-1}},\ n\ge 1.
$$
Then 
$$
T_n = \sum_{k=0}^{\lfloor n/2\rfloor} {n\choose 2k}(2k-1)!!.
$$
I will prove first @S.C.B.'s claim: 

Claim For each $n\ge 1$, $T_n$ and $T_{n+1}$ have no common odd prime divisor.

It relies on

Lemma For each $n\ge 1$, $T_n$ and $n$ have no common odd prime divisor.

Proof Let $p>2$ be an odd prime dividing $n$. Write
$$
T_n = 1 + \sum_{k=1}^{\lfloor n/2\rfloor} {n\choose 2k}(2k-1)!!.
$$
Note that $p\mid {n\choose 2k}$ for $1\le k< p/2$, and $p\mid (2k-1)!!$ for $k> p/2$. Therefore, $T_n = 1\pmod p$, yielding the statement.
Proof of the claim Let $p>2$ be a prime. Assume, by way of contradiction, that  $n$ is the smallest positive integer such that $p$ divides both $T_{n+1}$ and $T_n$. Then it follows from $(1)$ that $p\mid n$, contradicting Lemma.

Therefore, the only possibility for $x_n$ to be an integer is $T_{n-1}$ being a power of $2$. However, as it is explained in the Wikipedia page, this is not possible for $n\ge 4$.
A: This answer uses that for $n\ge 4$, 
$$\frac{1+\sqrt{4n-3}}{2}\lt x_n\lt \frac{1+\sqrt{4n+1}}{2}\tag1$$
The proof for $(1)$ is written at the end of this answer.
Multiplying $(1)$ by $2$, subtracting $1$ and squaring, we get
$$4n-3\lt 4(x_n^2-x_n)+1\lt 4n+1$$
Then, subtracting $1$ and dividing by $4$, we get
$$n-1\lt x_n^2-x_n\lt n$$
Since $x_n^2-x_n$ is not an integer, we have that $x_n$ is not an integer.
So, your conjecture is true.

Let us prove $(1)$ by induction.
For $n=4$, since $13\lt 16\lt 17$, we get
$$\frac{1+\sqrt{13}}{2}\lt \frac{1+\sqrt{16}}2=x_4\lt \frac{1+\sqrt{17}}{2}$$
Supposing that $(1)$ holds for some $n\ge 4$ gives
$$\begin{align}x_{n+1}-\frac{1+\sqrt{4(n+1)-3}}{2}&=1+\frac{n}{x_n}-\frac{1+\sqrt{4n+1}}{2}\\\\&\gt 1+\frac{2n}{1+\sqrt{4n+1}}-\frac{1+\sqrt{4n+1}}{2}=0\end{align}$$
and
$$\begin{align}\frac{1+\sqrt{4(n+1)+1}}{2}-x_{n+1}&=\frac{1+\sqrt{4n+5}}{2}-1-\frac{n}{x_n}\\\\&\gt \frac{1+\sqrt{4n+5}}{2}-1-\frac{2n}{1+\sqrt{4n-3}}\\\\&=\frac{1+\sqrt{4n+5}}{2}-1-\frac{2n(1-\sqrt{4n-3})}{1-(4n-3)}\\\\&=\frac{(n-1)\sqrt{4n+5}+1-n\sqrt{4n-3}}{2(n-1)}\\\\&=\frac{\sqrt{4n+5}-3}{(n-1)\sqrt{4n+5}+1+n\sqrt{4n-3}}\\\\&\gt 0\qquad\quad\blacksquare\end{align}$$
