Find all triplets $(a, k, m)$ of positive integers that satisfy the equation $k + a^k = m + 2a^m$.
To find all triplets \((a, k, m)\) of positive integers that satisfy the equation \(k + a^k = m + 2a^m\), we will proceed as follows:

1. **Rewrite the equation:**
   \[
   k + a^k = m + 2a^m
   \]
   This can be rearranged to:
   \[
   k - m = 2a^m - a^k
   \]

2. **Analyze the inequality:**
   If \(m \geq k\), then \(k - m \leq 0\) and \(2a^m - a^k \geq 2a^k - a^k = a^k\). This implies:
   \[
   k - m \leq 0 \quad \text{and} \quad 2a^m - a^k \geq a^k
   \]
   Since \(a^k\) is positive, this leads to a contradiction because \(k - m\) cannot be both non-positive and equal to a positive value. Therefore, \(m < k\).

3. **Divisibility condition:**
   Since \(m < k\), we can write:
   \[
   k - m = 2a^m - a^k
   \]
   This implies \(a^m\) divides \(k - m\). Let:
   \[
   k - m = c \cdot a^m \quad \text{for some integer } c
   \]
   Substituting back, we get:
   \[
   c \cdot a^m = 2a^m - a^k
   \]
   Simplifying, we have:
   \[
   c = 2 - a^{k-m}
   \]

4. **Analyze the exponent:**
   Since \(c\) must be a positive integer, we have:
   \[
   2 - a^{k-m} > 0 \implies a^{k-m} < 2
   \]
   The only positive integer \(a\) such that \(a^{k-m} < 2\) is \(a = 1\) (since \(k - m \geq 1\)).

5. **Substitute \(a = 1\):**
   If \(a = 1\), the original equation becomes:
   \[
   k + 1^k = m + 2 \cdot 1^m \implies k + 1 = m + 2
   \]
   Simplifying, we get:
   \[
   k + 1 = m + 2 \implies k = m + 1
   \]

6. **Conclusion:**
   Therefore, the only solutions in positive integers of the equation \(k + a^k = m + 2a^m\) are:
   \[
   (a, k, m) = (1, t+1, t) \quad \text{where } t \text{ is a positive integer}
   \]

The final answer is \( \boxed{ (a, k, m) = (1, t+1, t) } \) where \(t\) is a positive integer.