Find all $n$ natural numbers such that for each of them there exist $p , q$ primes such that these terms satisfy.

$1.$ $p+2=q$
$2.$ $2^n+p$ and $2^n+q$ are primes.
1. **Identify the relationship between \( p \) and \( q \):**
   Given \( p \) and \( q \) are primes such that \( p + 2 = q \). This implies \( q = p + 2 \).

2. **Conditions for \( 2^n + p \) and \( 2^n + q \) to be primes:**
   We need \( 2^n + p \) and \( 2^n + q \) to be primes.

3. **Consider the modulo 3 analysis:**
   Assume \( p \) and \( q \) are not divisible by 3. Then \( p \) and \( q \) must be \( 1 \) and \( 2 \pmod{3} \) (not necessarily in that order). Since \( q = p + 2 \), if \( p \equiv 1 \pmod{3} \), then \( q \equiv 3 \equiv 0 \pmod{3} \), which is a contradiction because \( q \) is prime and not equal to 3. Therefore, one of \( p \) or \( q \) must be 3.

4. **Check the case where \( q = 3 \):**
   If \( q = 3 \), then \( p = 1 \), but 1 is not a prime. Therefore, the only possible solution is \( p = 3 \) and \( q = 5 \).

5. **Verify \( n \) for \( p = 3 \) and \( q = 5 \):**
   We need \( 2^n + 3 \) and \( 2^n + 5 \) to be primes.

6. **Check small values of \( n \):**
   - For \( n = 1 \):
     \[
     2^1 + 3 = 5 \quad (\text{prime})
     \]
     \[
     2^1 + 5 = 7 \quad (\text{prime})
     \]
     Thus, \( n = 1 \) is a solution.

   - For \( n = 3 \):
     \[
     2^3 + 3 = 11 \quad (\text{prime})
     \]
     \[
     2^3 + 5 = 13 \quad (\text{prime})
     \]
     Thus, \( n = 3 \) is a solution.

7. **Prove that no \( n > 3 \) works:**
   Since \( 2^n \) must be \( 2 \pmod{3} \), \( n \) must be odd. Taking modulo 5, we get that \( n \) is not \( 1 \pmod{4} \) unless \( n = 1 \), which is already a solution. So assume \( n \equiv 3 \pmod{4} \).

8. **Consider modulo 7:**
   \( 2^n \) is not \( 4 \pmod{7} \), so \( n \ne 2 \pmod{3} \). Also, \( 2^n \) is not \( 2 \pmod{7} \), so \( n \ne 1 \pmod{3} \). Thus, \( 3 | n \). Therefore, \( n \) can be written as \( 3 \pmod{12} \).

9. **Consider modulo 13:**
   \( 13 | 2^n + 5 \). So \( 2^n + 5 = 13 \implies 2^n = 8 \implies n = 3 \), which is already a solution.

Thus, the only solutions are \( n = 1 \) and \( n = 3 \).

The final answer is \( \boxed{n = 1, 3} \).