Where's the missing $x$ ?

First let us consider the ranges of $y$ and $z$ . Note that $y$ is maximum when $x=z=1$ , then $y+y+1=10$ , $\implies 1 \le y \le 4$ . Similarly, $z$ is maximum, when $x=y=1$ and $1+z+z = 10$ , again $1 \le z \le 4$ . Then we have:

$\begin{aligned} xy + yz + z & = 10 \\ \implies x & = \frac {10-z}y - z \end{aligned}$

We note that $x$ is an integer only when $\dfrac {10-z}y$ is an integer or $y\mid 10-z$ . Therefore,

When $y=1$ , possible $z = \begin{cases} 1 \\ 2 \\3 \\ 4 \end{cases}$ and $\implies x = \begin{cases} 8 \\ 6 \\ 4 \\ 2 \end{cases}$ and solutions: $\begin{cases} (8,1,1) \\ (6,1,2) \\ (4,1,3) \\ (2,1,4) \end{cases}$

When $y=2$ , possible $z = \begin{cases} 2 \\ 4 \end{cases}$ and $\implies x = \begin{cases} 2 \\ \color{#D61F06} -1 \end{cases}$ and solution: $(2,2,2)$

When $y=3$ , possible $z = \begin{cases} 1 \\ 4 \end{cases}$ and $\implies x = \begin{cases} 2 \\ \color{#D61F06} -2 \end{cases}$ and solution: $(2,3,2)$

When $y=4$ , possible $z = 2$ and $\implies x = \color{#D61F06} 0$ ; there is no solution.

Therefore, there are $\boxed{6}$ solutions.

Noted that $x \geq 1$ and $y \geq 1$

So

$xy+yz+z\geq 1+z+z$

$2z+1\leq 10$

$z\leq 4.5$

$z=1,2,3,4$

Rewrite equation give us

$y=\frac{10-z}{x+z}$

When $z=1$ ,

$y=\frac{9}{x+1}$ leads to two answers $(2,3,1)$ , $(8,1,1)$ .

When $z=2$ ,

$y=\frac{8}{x+2}$ leads to two answers $(2,2,2)$ , $(6,1,2)$ .

When $z=3$ ,

$y=\frac{7}{x+3}$ leads to the only answer $(4,1,3)$ .

When $z=4$ ,

$y=\frac{6}{x+4}$ leads to the only answer $(2,1,4)$ .

So, there are $\boxed{6}$ solutions.