Determine the smallest integer $k > 1$ such that there exist $k$ distinct primes whose squares sum to a power of $2$.
1. We are given \( k > 1 \) distinct primes \( p_1 < p_2 < \cdots < p_k \) and an integer \( n \) such that 
   \[
   p_1^2 + p_2^2 + \cdots + p_k^2 = 2^n.
   \]
   Let \( S_k \) be the sum on the left-hand side of the equation. We need to find the smallest \( k \) such that \( S_k = 2^n \).

2. First, consider the smallest possible sum of squares of two distinct primes:
   \[
   2^2 + 3^2 = 4 + 9 = 13.
   \]
   Since \( 13 \) is not a power of \( 2 \), \( k \) must be greater than \( 2 \).

3. Next, consider the sum of squares of three distinct primes:
   \[
   2^2 + 3^2 + 5^2 = 4 + 9 + 25 = 38.
   \]
   Since \( 38 \) is not a power of \( 2 \), \( k \) must be greater than \( 3 \).

4. Now, consider the sum of squares of four distinct primes:
   \[
   2^2 + 3^2 + 5^2 + 7^2 = 4 + 9 + 25 + 49 = 87.
   \]
   Since \( 87 \) is not a power of \( 2 \), \( k \) must be greater than \( 4 \).

5. Consider the sum of squares of five distinct primes:
   \[
   2^2 + 3^2 + 5^2 + 7^2 + 11^2 = 4 + 9 + 25 + 49 + 121 = 208.
   \]
   Since \( 208 \) is not a power of \( 2 \), we need to check further.

6. We need to find a combination of five primes whose squares sum to a power of \( 2 \). Let's try replacing \( 11 \) with the next prime \( 13 \):
   \[
   2^2 + 3^2 + 5^2 + 7^2 + 13^2 = 4 + 9 + 25 + 49 + 169 = 256.
   \]
   Since \( 256 = 2^8 \), we have found a valid combination.

7. Therefore, the smallest \( k \) such that the sum of the squares of \( k \) distinct primes is a power of \( 2 \) is \( 5 \).

The final answer is \( \boxed{ k = 5 } \).