There are several different positive integers written on the blackboard, and the sum of any two different numbers should be should be a prime power. At this time, find the maximum possible number of integers written on the blackboard. A prime power is an integer expressed in the form $p^n$ using a prime number $p$ and a non-negative integer number $n$.
1. **Lemma**: You cannot have three odd integers at the same time. 

   **Proof**: Suppose \( a > b > c \) are three odd integers. We must have \( a + b = 2^x \), \( a + c = 2^y \), and \( b + c = 2^z \) with \( x \ge y \ge z \ge 2 \). Adding the first two equations gives:
   \[
   2a + b + c = 2^x + 2^y
   \]
   Since \( b + c = 2^z \), we can substitute to get:
   \[
   2a + 2^z = 2^x + 2^y
   \]
   This implies:
   \[
   2a = 2^x + 2^y - 2^z
   \]
   Dividing by 2, we get:
   \[
   a = 2^{x-1} + 2^{y-1} - 2^{z-1}
   \]
   Since \( a \) is odd, the right-hand side must be odd. However, \( 2^{x-1} + 2^{y-1} - 2^{z-1} \) is even, leading to a contradiction. Therefore, we cannot have three odd integers at the same time. \(\blacksquare\)

2. **Observation**: We cannot have three even integers at the same time.

   Suppose our even integers are \( 2^a x, 2^b y, 2^c z \), with \( x, y, z \) being odd integers. We must have \( a = b = c \) because if two aren't equal (suppose WLOG \( a \) and \( b \), and \( a > b \)), then we have:
   \[
   2^a x + 2^b y = 2^b (2^{a-b} x + y) = 2^k \implies 2^{a-b} x + y = 2^{k-b}
   \]
   Since \( x \) and \( y \) are odd, \( 2^{a-b} x + y \) is odd, which cannot equal \( 2^{k-b} \) (an even number). Therefore, \( a = b = c \).

3. **Combining the observations**: We can have at most 2 even and 2 odd integers. 

   To find a construction, consider the integers \( 2, 3, 5, 6 \). We check the sums:
   \[
   2 + 3 = 5 \quad (\text{prime})
   \]
   \[
   2 + 5 = 7 \quad (\text{prime})
   \]
   \[
   2 + 6 = 8 \quad (2^3)
   \]
   \[
   3 + 5 = 8 \quad (2^3)
   \]
   \[
   3 + 6 = 9 \quad (3^2)
   \]
   \[
   5 + 6 = 11 \quad (\text{prime})
   \]
   All sums are prime powers, confirming that the construction is valid.

The final answer is \(\boxed{4}\).