For natural numbers $x$ and $y$, let $(x,y)$ denote the greatest common divisor of $x$ and $y$. How many pairs of natural numbers $x$ and $y$ with $x \le y$ satisfy the equation $xy = x + y + (x, y)$?
1. Let \( x = ha \) and \( y = hb \) for some natural numbers \( h, a, \) and \( b \). Notice that \(\gcd(x, y) = h\).
2. Substitute \( x = ha \) and \( y = hb \) into the given equation \( xy = x + y + \gcd(x, y) \):
   \[
   (ha)(hb) = ha + hb + h
   \]
3. Simplify the equation:
   \[
   h^2 ab = h(a + b) + h
   \]
4. Factor out \( h \) from the right-hand side:
   \[
   h^2 ab = h(a + b + 1)
   \]
5. Divide both sides by \( h \) (assuming \( h \neq 0 \)):
   \[
   hab = a + b + 1
   \]
6. Rearrange the equation to solve for \( h \):
   \[
   h = \frac{a + b + 1}{ab}
   \]
7. Since \( h \) must be a natural number, \( ab \) must divide \( a + b + 1 \). Therefore, we have:
   \[
   a + b + 1 \geq ab
   \]
8. Rearrange the inequality:
   \[
   ab - a - b \leq 1
   \]
9. Add 1 to both sides:
   \[
   ab - a - b + 1 \leq 2
   \]
10. Factor the left-hand side:
    \[
    (a - 1)(b - 1) \leq 2
    \]
11. Consider the possible values for \((a - 1)(b - 1)\):
    - If \((a - 1)(b - 1) = 0\), then either \(a = 1\) or \(b = 1\).
    - If \((a - 1)(b - 1) = 1\), then \((a - 1, b - 1)\) could be \((1, 1)\), so \(a = 2\) and \(b = 2\).
    - If \((a - 1)(b - 1) = 2\), then \((a - 1, b - 1)\) could be \((1, 2)\) or \((2, 1)\), so \((a, b)\) could be \((2, 3)\) or \((3, 2)\).

12. List the possible pairs \((a, b)\):
    - \((1, 1)\)
    - \((2, 3)\)
    - \((3, 2)\)

13. Convert these pairs back to \((x, y)\) using \( x = ha \) and \( y = hb \):
    - For \((a, b) = (1, 1)\), \( h = \frac{1 + 1 + 1}{1 \cdot 1} = 3 \), so \((x, y) = (3, 3)\).
    - For \((a, b) = (2, 3)\), \( h = \frac{2 + 3 + 1}{2 \cdot 3} = 1 \), so \((x, y) = (2, 3)\).
    - For \((a, b) = (3, 2)\), \( h = \frac{3 + 2 + 1}{3 \cdot 2} = 1 \), so \((x, y) = (3, 2)\).

14. Since \( x \leq y \), the valid pairs are:
    - \((3, 3)\)
    - \((2, 3)\)
    - \((3, 2)\) (not valid since \( x \leq y \))

15. Therefore, the number of valid pairs is 3.

The final answer is \(\boxed{3}\)