How many ordered integer pairs $(x,y)$ ($0\leq x,y < 31$) are there satisfying $(x^2-18)^2\equiv y^2 (\mod 31)$?

$ \textbf{(A)}\ 59
\qquad\textbf{(B)}\ 60
\qquad\textbf{(C)}\ 61
\qquad\textbf{(D)}\ 62
\qquad\textbf{(E)}\ \text{None}
$
1. We start with the given congruence:
   \[
   (x^2 - 18)^2 \equiv y^2 \pmod{31}
   \]
   Let \( A = x^2 - 18 \). Then the equation becomes:
   \[
   A^2 \equiv y^2 \pmod{31}
   \]

2. This congruence implies:
   \[
   (A - y)(A + y) \equiv 0 \pmod{31}
   \]
   Therefore, \( A \equiv y \pmod{31} \) or \( A \equiv -y \pmod{31} \).

3. For each \( x \) in the range \( 0 \leq x < 31 \), we need to determine the possible values of \( y \). Since \( A = x^2 - 18 \), we have:
   \[
   y \equiv x^2 - 18 \pmod{31} \quad \text{or} \quad y \equiv -(x^2 - 18) \pmod{31}
   \]

4. This means for each \( x \), there are two possible values of \( y \):
   \[
   y_1 = x^2 - 18 \pmod{31} \quad \text{and} \quad y_2 = -(x^2 - 18) \pmod{31}
   \]

5. Since \( x \) ranges from 0 to 30 (inclusive), there are 31 possible values for \( x \). For each \( x \), there are 2 corresponding values of \( y \), giving us:
   \[
   31 \times 2 = 62
   \]

6. However, we need to consider the special case where \( y \equiv 0 \pmod{31} \). This happens when:
   \[
   x^2 - 18 \equiv 0 \pmod{31} \quad \Rightarrow \quad x^2 \equiv 18 \pmod{31}
   \]

7. We need to find the values of \( x \) such that \( x^2 \equiv 18 \pmod{31} \). By checking the quadratic residues modulo 31, we find that:
   \[
   x \equiv \pm 7 \pmod{31}
   \]
   are the solutions to \( x^2 \equiv 18 \pmod{31} \).

8. Therefore, there are 2 values of \( x \) (namely \( x = 7 \) and \( x = 24 \)) that make \( y \equiv 0 \pmod{31} \). For these values of \( x \), there is only one corresponding value of \( y \) instead of two.

9. Thus, we need to subtract 1 for each of these special cases from the total count:
   \[
   62 - 2 = 60
   \]

Conclusion:
The number of ordered integer pairs \((x, y)\) satisfying the given congruence is \(\boxed{60}\).