Is there any solution such that.. Is there any positive integer solutions of $x$, $y$, $z$ such that both $$\frac{x^2+y^2+z^2}{x+y+z}$$ and $$\frac{x^2+y^2+z^2}{xyz}$$ are integers?

Edit: Forgot to mention that $x$, $y$, $z$ are distinct positive integers.
A remark, how to start: The second condition can be written as Diophantine equation 
$$
x^2+y^2+z^2=nxyz,
$$
and it is easy to see that we must have $n=1$ or $n=3$. This is already very helpful. Compare also with this question.

For $n=3$ the triples $(x,y,z)$ with $\frac{x^2+y^2+z^2}{xyz}=3$ are known as Markov numbers, and we can produce a Markov tree. Now it gets easier if we also require that $(x,y,z)$ is a solution of the Diophantine equation $x^2+y^2+z^2=m(x+y+z)$.
Special case

Well, if $x = y = z$ then your equations become

$$\frac{3x^2}{3x} = x$$

$$\frac{3x^2}{x^3} = \frac{3}{x}$$

If they are coupled, then you notice that $x = 1$ is the simplest result. There is also $x = 3$: $3/x$ is an integer if and only if $x = 1$ or $x = 3$.

Unless you count negative integers too, in that case you have also $x = y = z = -1$ and $x = y = z = -3$