How many real solutions does the following system have? $\begin{cases} x+y=2  \\
 xy - z^2 = 1 \end{cases}$
To determine the number of real solutions for the given system of equations:
\[
\begin{cases}
x + y = 2 \\
xy - z^2 = 1
\end{cases}
\]

1. **Express \( y \) in terms of \( x \) from the first equation:**
   \[
   y = 2 - x
   \]

2. **Substitute \( y = 2 - x \) into the second equation:**
   \[
   x(2 - x) - z^2 = 1
   \]

3. **Simplify the equation:**
   \[
   2x - x^2 - z^2 = 1
   \]

4. **Rearrange the equation:**
   \[
   -x^2 + 2x - z^2 = 1
   \]

5. **Move all terms to one side to form a standard quadratic equation:**
   \[
   x^2 - 2x + z^2 = 0
   \]

6. **Complete the square for the \( x \)-terms:**
   \[
   (x - 1)^2 + z^2 = 1
   \]

7. **Analyze the equation \((x - 1)^2 + z^2 = 1\):**
   This represents a circle in the \( xz \)-plane with center at \( (1, 0) \) and radius 1.

8. **Determine the points on the circle:**
   Since \((x - 1)^2 + z^2 = 1\) is a circle, we need to find the points \((x, z)\) that satisfy this equation. For each \( x \) value within the interval \([0, 2]\), there is a corresponding \( z \) value such that the equation holds true.

9. **Verify the corresponding \( y \) values:**
   For each \( x \) value, \( y = 2 - x \). Since \( x \) ranges from 0 to 2, \( y \) will also range from 0 to 2.

10. **Count the number of real solutions:**
    Since the circle equation \((x - 1)^2 + z^2 = 1\) has infinitely many points on it, and for each point \((x, z)\), there is a corresponding \( y \) value, the system has infinitely many real solutions.

The final answer is \(\boxed{\text{infinitely many}}\).