Real Solutions

$xy = z^2 + 1 > 0$

$x + y = 2 > 0 \therefore x , y > 0$

By AM - GM Inequality ,

$x + y \geq 2\cdot (xy)^{\frac{1}{2}}$

$\Rightarrow 1 \geq xy = 1 + z^2 \geq 1$

$\therefore 1 + z^2 = 1 \Rightarrow z = 0 , x = y = 1$

Hence , one triplet $\boxed{(x , y , z) = (1 , 1 , 0)}$

Substitute $y = 2 - x$ into the second equation to get

$x(2 - x) - z^{2} = 1 \Longrightarrow -(x^{2} - 2x + 1) + 1 = 1 + z^{2} \Longrightarrow -(x - 1)^{2} = z^{2}$ .

Since squares of real numbers are necessarily non-negative, the only way for this equation to hold is to have

$x - 1 = 0 \Longrightarrow x = 1, y = 2 - x = 1$ and $z = 0$ .

Thus there is just $\boxed{1}$ triplet that satisfies the given system of equations, namely $(1,1,0)$ .

An alternative solution (to AM - GM), by using quadratics:

$x + y = 2$ $xy - z^2 = 1$

$xy = z^2 + 1$

From the latter equation, we can see that if z is a real number, then xy≥1 and this also means, that x ≠ 0 and y ≠ 0.

$y = \frac {z^2+1}{x}$

After substituting this into our first equation, we get:

$x + \frac {z^2+1}{x} = 2$

$x^2 - 2x + z^2+1 = 0$

Applying the quadratic formula:

$x = \frac { 2 ± \sqrt {2^2 - 4×1×(z^2 + 1)} }{2×1} = \frac { 2 ± \sqrt {- 4z^2 } }{2} = 1 ± \sqrt {- z^2 }$ .... (I)

Our result for x is a real number iff the expression under the square root (the simplified discriminant) is non-negative:

$- z^2 ≥ 0 \iff z^2 < 0 \iff z = 0$

This gives us the only solution: x = y = 1, z = 0.

$\text {Hence, our answer should be: } \boxed {1}$

Remark:

$\text {If we allow } z \in \mathbb {C} \text { while } x, y \in \mathbb {R} \text { , then we have infinitely many solutions: }$

$\text { (I) gives us: } z = \pm ai , x = 1 \pm a , y = \frac {1 - a^2}{1 \pm a} = 1 \mp a$

$\text { where } a \in \mathbb {R}^+ , a ≠ 1$

We will also get (x, y, z) solutions in the xy = 0 case:

(0, 2, i), (0, 2, -i), (2, 0, i) and (0, 2, i).

Consider rectangular hyperbola: $xy =1 + z^{2}$

Aim is to find points of intersection between the hyperbola and $x + y =2$

The line is not present in the 4th quadrant. Therefore, by AM GM $x=1, y=1, z=0$

if z > 0, xy > 1 No point of intersection.