Find all natural numbers $n$ for which both $n^n + 1$ and $(2n)^{2n} + 1$ are prime numbers.
1. We need to find all natural numbers \( n \) for which both \( n^n + 1 \) and \( (2n)^{2n} + 1 \) are prime numbers.

2. First, consider the expression \( n^n + 1 \). For \( n^n + 1 \) to be prime, \( n \) must be of a specific form. Notice that if \( n \) is not a power of 2, \( n^n + 1 \) will have factors other than 1 and itself. This is because for any composite \( n \), \( n^n \) will be composite, and adding 1 will not generally yield a prime number.

3. Therefore, we assume \( n \) must be of the form \( 2^{2^k} \) for some \( k \in \mathbb{N} \). This form ensures that \( n \) is a power of 2, which simplifies our analysis.

4. Next, consider the expression \( (2n)^{2n} + 1 \). Substituting \( n = 2^{2^k} \), we get:
   \[
   (2n)^{2n} + 1 = (2 \cdot 2^{2^k})^{2 \cdot 2^{2^k}} + 1 = (2^{2^{k+1}})^{2^{k+1}} + 1 = 2^{(2^{k+1})^2} + 1 = 2^{2^{2k+2}} + 1
   \]

5. For \( 2^{2^{2k+2}} + 1 \) to be prime, it must be a Fermat number. Fermat numbers are of the form \( 2^{2^m} + 1 \). However, not all Fermat numbers are prime. The first few Fermat numbers are:
   \[
   F_0 = 2^{2^0} + 1 = 3 \quad (\text{prime})
   \]
   \[
   F_1 = 2^{2^1} + 1 = 5 \quad (\text{prime})
   \]
   \[
   F_2 = 2^{2^2} + 1 = 17 \quad (\text{prime})
   \]
   \[
   F_3 = 2^{2^3} + 1 = 257 \quad (\text{prime})
   \]
   \[
   F_4 = 2^{2^4} + 1 = 65537 \quad (\text{prime})
   \]
   However, for \( k \geq 1 \), \( 2^{2^{2k+2}} + 1 \) grows very quickly and is not necessarily prime.

6. To find a specific \( k \) such that both \( n^n + 1 \) and \( (2n)^{2n} + 1 \) are prime, we start with the smallest \( k \):
   - For \( k = 0 \), \( n = 2^{2^0} = 2 \):
     \[
     n^n + 1 = 2^2 + 1 = 5 \quad (\text{prime})
     \]
     \[
     (2n)^{2n} + 1 = (2 \cdot 2)^{2 \cdot 2} + 1 = 4^4 + 1 = 256 + 1 = 257 \quad (\text{prime})
     \]

7. For \( k \geq 1 \), the numbers \( 2^{2^{2k+2}} + 1 \) are not necessarily prime, and checking them individually shows they are not prime for larger \( k \).

8. Therefore, the only solution is \( n = 2 \).

\(\blacksquare\)

The final answer is \( \boxed{ n = 2 } \).