Counting integral solutions

Because x, y, z are positive integers, we can divide both sides by y^2 and, defining a as a= x/y, move all terms to the left to get:

$a^2-2az-1=0$

$a^2-2az+z^2=z^2+1$

$(a-z)^2=z^2+1$

which is impossible because a is rational and z is a positive integer, leading to z^2+1 not being the square of any rational number. Thus, there are $\boxed{0}$ such triplets.

Assume that the positive integers $x,y,z$ satisfy the given equation and let $d=xy$ . If $d=1$ then $x=y=1$ and $z=0$ which cannot happen.

Hence $d>1$ . Let $p$ be a prime divisor of $d$

Now $(x+y)(x-y)=x^2-y^2=2xyz \equiv 0 \pmod{p}$

Either $x \equiv y \pmod{p}$ or $x \equiv -y \pmod{p}$

But $p$ divides one of $x$ and $y$ , so $p$ must divide the other too

Hence $x_1=\frac{x}{p}$ and $y_1=\frac{y}{p}$ are positive integers and $x_1,y_1,z$ satisfy the given equation as well.

We can construct an infinte sequence of solutions $(x_n,y_n,z),n \geq 1$ to the original equation with $x_1>x_2>x_3 >...$

By Fermat's method of infinite descent this is impossible

Hence the given equation has no integral solutions in $x,y,z$