How many pairs of real numbers $ \left(x,y\right)$ are there such that $ x^{4} \minus{} 2^{ \minus{} y^{2} } x^{2} \minus{} \left\| x^{2} \right\| \plus{} 1 \equal{} 0$, where $ \left\| a\right\|$ denotes the greatest integer not exceeding $ a$.

$\textbf{(A)}\ 0   \qquad\textbf{(B)}\ 1   \qquad\textbf{(C)}\ 2   \qquad\textbf{(D)}\ 4   \qquad\textbf{(E)}\ \text{Infinitely many}$
1. **Rewrite the given equation:**
   \[
   x^4 - 2^{-y^2} x^2 - \left\| x^2 \right\| + 1 = 0
   \]
   We can rewrite this equation by letting \( z = x^2 \). Thus, the equation becomes:
   \[
   z^2 - 2^{-y^2} z - \left\| z \right\| + 1 = 0
   \]

2. **Analyze the equation:**
   Since \( z = x^2 \), \( z \geq 0 \). Also, \( \left\| z \right\| \) is the greatest integer not exceeding \( z \). Therefore, \( \left\| z \right\| \leq z < \left\| z \right\| + 1 \).

3. **Consider the possible values of \( z \):**
   - If \( z \) is an integer, then \( \left\| z \right\| = z \).
   - If \( z \) is not an integer, then \( \left\| z \right\| \) is the integer part of \( z \).

4. **Case 1: \( z \) is an integer:**
   If \( z \) is an integer, then \( \left\| z \right\| = z \). The equation simplifies to:
   \[
   z^2 - 2^{-y^2} z - z + 1 = 0
   \]
   \[
   z^2 - (1 + 2^{-y^2}) z + 1 = 0
   \]

5. **Solve for \( z \):**
   For the quadratic equation \( z^2 - (1 + 2^{-y^2}) z + 1 = 0 \) to have real solutions, the discriminant must be non-negative:
   \[
   \Delta = (1 + 2^{-y^2})^2 - 4 \cdot 1 \cdot 1 \geq 0
   \]
   \[
   (1 + 2^{-y^2})^2 - 4 \geq 0
   \]
   \[
   1 + 2 \cdot 2^{-y^2} + 2^{-2y^2} - 4 \geq 0
   \]
   \[
   2 \cdot 2^{-y^2} + 2^{-2y^2} - 3 \geq 0
   \]

6. **Simplify the inequality:**
   Let \( t = 2^{-y^2} \). Then the inequality becomes:
   \[
   2t + t^2 - 3 \geq 0
   \]
   Solve the quadratic inequality:
   \[
   t^2 + 2t - 3 \geq 0
   \]
   \[
   (t + 3)(t - 1) \geq 0
   \]
   The solutions to this inequality are:
   \[
   t \leq -3 \quad \text{or} \quad t \geq 1
   \]
   Since \( t = 2^{-y^2} \) and \( 2^{-y^2} \) is always positive, we have:
   \[
   2^{-y^2} \geq 1 \implies y^2 = 0 \implies y = 0
   \]

7. **Substitute \( y = 0 \) back into the equation:**
   If \( y = 0 \), then \( 2^{-y^2} = 1 \). The equation becomes:
   \[
   z^2 - 2z + 1 = 0
   \]
   \[
   (z - 1)^2 = 0
   \]
   \[
   z = 1
   \]
   Since \( z = x^2 \), we have \( x^2 = 1 \), so \( x = \pm 1 \).

8. **Conclusion:**
   The pairs \((x, y)\) that satisfy the equation are \((1, 0)\) and \((-1, 0)\). Therefore, there are 2 pairs of real numbers \((x, y)\) that satisfy the given equation.

The final answer is \(\boxed{2}\)