Suppose $p < q < r < s$ are prime numbers such that $pqrs + 1 = 4^{p+q}$. Find $r + s$.
1. **Identify the properties of the primes:**
   Given that \( p < q < r < s \) are prime numbers and \( pqrs + 1 = 4^{p+q} \), we first note that \( p, q, r, s \) cannot be 2 because if any of them were 2, the left-hand side (LHS) would be odd, while the right-hand side (RHS) would be even, which is a contradiction. Therefore, \( p, q, r, s \) are all odd primes.

2. **Analyze the parity of \( p+q \):**
   Since \( p \) and \( q \) are odd primes, their sum \( p+q \) is even. Let \( p+q = 2k \) for some integer \( k \). This implies that \( 4^{p+q} = 4^{2k} = (2^2)^{2k} = 2^{4k} \).

3. **Use modular arithmetic:**
   We know that \( 4^{p+q} - 1 \) must be divisible by \( 15 \) because \( 4^2 - 1 = 15 \). Therefore, \( 4^{p+q} - 1 \equiv 0 \pmod{15} \). This implies that \( 4^{2k} \equiv 1 \pmod{15} \).

4. **Determine \( p \) and \( q \):**
   The smallest primes \( p \) and \( q \) that satisfy \( 4^{p+q} - 1 \equiv 0 \pmod{15} \) are \( p = 3 \) and \( q = 5 \). This is because \( 4^{3+5} = 4^8 \equiv 1 \pmod{15} \).

5. **Calculate \( 4^{p+q} - 1 \):**
   With \( p = 3 \) and \( q = 5 \), we have:
   \[
   4^{p+q} = 4^8 = 65536
   \]
   Therefore,
   \[
   4^{p+q} - 1 = 65536 - 1 = 65535
   \]

6. **Solve for \( r \) and \( s \):**
   Given \( pqrs + 1 = 65536 \), we have:
   \[
   pqrs = 65535
   \]
   Substituting \( p = 3 \) and \( q = 5 \), we get:
   \[
   3 \cdot 5 \cdot r \cdot s = 65535
   \]
   Simplifying, we find:
   \[
   15rs = 65535 \implies rs = \frac{65535}{15} = 4369
   \]

7. **Factorize 4369:**
   We need to find the prime factors of 4369. Checking for divisibility, we find:
   \[
   4369 = 17 \times 257
   \]
   Both 17 and 257 are prime numbers.

8. **Sum \( r \) and \( s \):**
   Therefore, \( r = 17 \) and \( s = 257 \). Summing these, we get:
   \[
   r + s = 17 + 257 = 274
   \]

The final answer is \( \boxed{274} \).