Find the smallest positive integer $n \geq 10$ such that $n + 6$ is a prime and $9n + 7$ is a perfect square.
1. Let \( 9n + 7 = x^2 \) for some \( x \in \mathbb{N} \). This implies:
   \[
   x^2 \equiv 7 \pmod{9}
   \]
2. We need to find \( x \) such that \( x^2 \equiv 7 \pmod{9} \). By checking the quadratic residues modulo 9, we find:
   \[
   x \equiv \pm 4 \pmod{9}
   \]
   Therefore, \( x = 9m \pm 4 \) for some \( m \in \mathbb{N} \).

3. Substitute \( x = 9m \pm 4 \) into the equation \( 9n + 7 = x^2 \):
   \[
   9n + 7 = (9m \pm 4)^2
   \]
   Expanding the square, we get:
   \[
   9n + 7 = 81m^2 \pm 72m + 16
   \]
4. Simplify the equation:
   \[
   9n + 7 = 81m^2 \pm 72m + 16 \implies 9n = 81m^2 \pm 72m + 9 \implies n = 9m^2 \pm 8m + 1
   \]

5. We need \( n + 6 \) to be a prime number:
   \[
   n + 6 = 9m^2 \pm 8m + 7
   \]

6. Consider the case \( n + 6 = 9m^2 + 8m + 7 \):
   - For \( m = 1 \):
     \[
     n + 6 = 9(1)^2 + 8(1) + 7 = 24 \quad (\text{not prime})
     \]
   - For \( m = 2 \):
     \[
     n + 6 = 9(2)^2 + 8(2) + 7 = 59 \quad (\text{prime})
     \]
     Therefore, \( n = 59 - 6 = 53 \).

7. Consider the case \( n + 6 = 9m^2 - 8m + 7 \):
   - For \( m = 1 \):
     \[
     n + 6 = 9(1)^2 - 8(1) + 7 = 8 \quad (\text{not prime})
     \]
   - For \( m = 2 \):
     \[
     n + 6 = 9(2)^2 - 8(2) + 7 = 15 \quad (\text{not prime})
     \]
   - For \( m \geq 3 \):
     \[
     n + 6 = 9m^2 - 8m + 7 > 59 \quad (\text{not minimum})
     \]

8. The smallest \( n \geq 10 \) such that \( n + 6 \) is a prime and \( 9n + 7 \) is a perfect square is \( n = 53 \).

The final answer is \( \boxed{53} \)