How many pairs of positive integers $(x,y)$ are there such that $\sqrt{xy}-71\sqrt x + 30 = 0$?

$ 
\textbf{(A)}\ 8
\qquad\textbf{(B)}\ 18
\qquad\textbf{(C)}\ 72
\qquad\textbf{(D)}\ 2130
\qquad\textbf{(E)}\ \text{Infinitely many}
$
1. Given the equation:
   \[
   \sqrt{xy} - 71\sqrt{x} + 30 = 0
   \]
   Let's set \( \sqrt{x} = a \) and \( \sqrt{y} = b \). Then the equation becomes:
   \[
   ab - 71a + 30 = 0
   \]

2. We can factor out \( a \) from the equation:
   \[
   a(b - 71) + 30 = 0
   \]
   Rearranging, we get:
   \[
   a(b - 71) = -30
   \]

3. Since \( a \) and \( b \) must be positive integers, we need to find pairs \((a, b)\) such that:
   \[
   a(b - 71) = -30
   \]
   This implies:
   \[
   b - 71 = -\frac{30}{a}
   \]
   Therefore:
   \[
   b = 71 - \frac{30}{a}
   \]

4. For \( b \) to be an integer, \( \frac{30}{a} \) must be an integer. Thus, \( a \) must be a divisor of 30. The positive divisors of 30 are:
   \[
   1, 2, 3, 5, 6, 10, 15, 30
   \]

5. We now check each divisor to find the corresponding \( b \):
   - For \( a = 1 \):
     \[
     b = 71 - \frac{30}{1} = 71 - 30 = 41
     \]
   - For \( a = 2 \):
     \[
     b = 71 - \frac{30}{2} = 71 - 15 = 56
     \]
   - For \( a = 3 \):
     \[
     b = 71 - \frac{30}{3} = 71 - 10 = 61
     \]
   - For \( a = 5 \):
     \[
     b = 71 - \frac{30}{5} = 71 - 6 = 65
     \]
   - For \( a = 6 \):
     \[
     b = 71 - \frac{30}{6} = 71 - 5 = 66
     \]
   - For \( a = 10 \):
     \[
     b = 71 - \frac{30}{10} = 71 - 3 = 68
     \]
   - For \( a = 15 \):
     \[
     b = 71 - \frac{30}{15} = 71 - 2 = 69
     \]
   - For \( a = 30 \):
     \[
     b = 71 - \frac{30}{30} = 71 - 1 = 70
     \]

6. Each pair \((a, b)\) corresponds to a pair \((\sqrt{x}, \sqrt{y})\). Since \( a \) and \( b \) are integers, \( x \) and \( y \) are perfect squares. Therefore, there are 8 pairs of positive integers \((x, y)\) that satisfy the given equation.

The final answer is \(\boxed{8}\)