Find all pairs of $ (a, n) $ natural numbers such that $ \varphi (a ^ n + n) = 2 ^ n. $ 
($ \varphi (n) $ is the Euler function, that is, the number of integers from $1$ up to $ n $, relative prime to $ n $)
1. **Understanding the problem**: We need to find all pairs of natural numbers \((a, n)\) such that \(\varphi(a^n + n) = 2^n\). Here, \(\varphi\) is the Euler's totient function, which counts the number of integers up to \(n\) that are relatively prime to \(n\).

2. **Prime divisors of \(a^n + n\)**: Since \(\varphi(a^n + n) = 2^n\), the number \(a^n + n\) must have only prime divisors that are either 2 or Fermat primes (primes of the form \(2^{2^k} + 1\)).

3. **Expression for \(\varphi(a^n + n)\)**: We can write:
   \[
   \frac{\varphi(a^n + n)}{a^n + n} = \prod_{p \in S} \frac{p-1}{p}
   \]
   where \(S\) is the set of prime divisors of \(a^n + n\).

4. **Lower bound of the product**: The right-hand side (RHS) of the above equation is at least:
   \[
   \prod_{i=1}^{\infty} \frac{2^{2^i}}{2^{2^i} + 1}
   \]
   We claim that:
   \[
   \frac{2^{2^i}}{2^{2^i} + 1} \ge \left(\frac{2^{2^{i-1}}}{2^{2^{i-1}} + 1}\right)^{\frac{2}{3}}
   \]
   Cubing and expanding, we want:
   \[
   2^{3 \cdot 2^i} \cdot (2^{2^{i-1}} + 1)^2 \ge 2^{2^i} \cdot (2^{2^i} + 1)^3
   \]
   This inequality can be verified.

5. **Estimate of the product**: Using the above estimate, it is clear that the RHS of the product is at least \(\frac{1}{8}\). Therefore:
   \[
   \frac{2^n}{a^n + n} \ge \frac{1}{8}
   \]
   This implies:
   \[
   2^n \ge \frac{a^n + n}{8}
   \]
   Hence, if \(a > 2\), we get \(n < 6\).

6. **Case analysis**:
   - If \(a = 1\), then \(a^n + n = 1^n + n = n + 1\). We need \(\varphi(n + 1) = 2^n\). However, \(n + 1 > 2^n\) for all \(n \ge 1\), which is a contradiction.
   - If \(a = 2\), we need to solve \(\varphi(2^n + n) = 2^n\).

7. **Checking small values of \(n\)**:
   - For \(n = 1\), \(2^1 + 1 = 3\), \(\varphi(3) = 2\), which is \(2^1\). So, \((a, n) = (2, 1)\) is a solution.
   - For \(n = 2\), \(2^2 + 2 = 6\), \(\varphi(6) = 2\), which is not \(2^2\).
   - For \(n = 3\), \(2^3 + 3 = 11\), \(\varphi(11) = 10\), which is not \(2^3\).
   - For \(n = 4\), \(2^4 + 4 = 20\), \(\varphi(20) = 8\), which is not \(2^4\).
   - For \(n = 5\), \(2^5 + 5 = 37\), \(\varphi(37) = 36\), which is not \(2^5\).

8. **Conclusion**: The only solution is \((a, n) = (2, 1)\).

The final answer is \( \boxed{ (2, 1) } \)