Natural number solutions

From the AM-GM inequality,

$\dfrac{5}{2} = \dfrac{x}{y} + \dfrac{y}{z+1} + \dfrac{z}{x} \geq 3 \sqrt[3]{\dfrac{x}{y} \left ( \dfrac{y}{z+1} \right ) \left ( \dfrac{z}{x} \right )} = 3 \sqrt[3]{\dfrac{z}{z+1}}.$

Thus, $\dfrac{z}{z+1} \leq \dfrac{125}{216}.$ The only positive integer solution to this inequality is $z = 1.$

We can plug this value of $z$ back into our original equation to get

$\dfrac{x}{y} + \dfrac{y}{2} + \dfrac{1}{x} = \dfrac{5}{2}.$

Note that we must have $y \leq 4,$ because if $y \geq 5,$ then we have $\dfrac{x}{y} + \dfrac{1}{x} = \dfrac{5 - y}{2} \leq 0,$ which is a contradiction as $x$ and $y$ is positive. Thus, we have 4 cases, which are all fairly simple to deal with:

If $y = 1,$ then $x + \dfrac{1}{x} = 2,$ which has solution $x = 1.$ Thus, $(x, y, z) = (1, 1, 1).$

If $y = 2,$ then $\dfrac{x}{2} + \dfrac{1}{x} = \dfrac{3}{2},$ which has solutions $x = 1, 2.$ Thus, $(x, y, z) = (1, 2, 1), (2, 2, 1).$

If $y = 3,$ then $\dfrac{x}{3} + \dfrac{1}{x} = 1,$ which has no positive integer solutions.

If $y = 4,$ then $\dfrac{x}{4} + \dfrac{1}{x} = \dfrac{1}{2},$ which has no positive integer solutions.

Therefore, there are $\boxed{3}$ total solutions in positive integers to the given equation: $(x, y, z) = (1, 1, 1), (1, 2, 1), (2, 2, 1).$

Check:

$\begin{aligned} (x, y, z) = (1, 1, 1) &\longrightarrow \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{1} = \dfrac{5}{2}\,\, \checkmark \\ (x, y, z) = (1, 2, 1) &\longrightarrow \dfrac{1}{2} + \dfrac{2}{2} + \dfrac{1}{1} = \dfrac{5}{2}\,\, \checkmark \\ (x, y, z) = (2, 2, 1) &\longrightarrow \dfrac{2}{2} + \dfrac{2}{2} + \dfrac{1}{2} = \dfrac{5}{2}\,\, \checkmark \end{aligned}$

By AM-GM we obtain that z must be equal to 1

\sqrt[3]{\dfrac{x}{y} \cdot \dfrac{y}{z+1} \cdot \dfrac{z}{x}} \leq \dfrac{\dfrac{x}{y} +\dfrac{y}{z+1} +\dfrac{z}{x}}{3} = \dfrac{5}{6}

$\dfrac{z}{z+1}\leq (5/6)^{3}\approx 0.5787$

$\dfrac{z}{z+1}$ is smaller or equal to $0.5787$ only for $z=1$

Our new diophantine equation becames:

$\dfrac{x}{y} + \dfrac{y}{2} + \dfrac{1}{x}=\dfrac{5}{2}$

Now we can express y in function of x:

$y=\dfrac{5-\dfrac{2}{x}\pm \sqrt{\dfrac{-8x^{3}+25x^{2}-20x+4}{x^{2}}}}{2}$

The function under the square root is strictly decreasing and assume a non-negative value only for x=1 or x=2

It is easy now finding the solutions and they are: $\boxed{ (1,1,1,) (1;2;1) (2,2,1) }$