Find all intengers n such that $2^n - 1$ is a multiple of 3 and $(2^n - 1)/3$ is a divisor of $4m^2 + 1$ for some intenger m.
To solve the problem, we need to find all integers \( n \) such that \( 2^n - 1 \) is a multiple of 3 and \( \frac{2^n - 1}{3} \) is a divisor of \( 4m^2 + 1 \) for some integer \( m \).

1. **Condition Analysis:**
   - First, we need \( 2^n - 1 \) to be a multiple of 3. This implies \( 2^n \equiv 1 \pmod{3} \). Since \( 2^2 \equiv 1 \pmod{3} \), \( n \) must be even. Therefore, let \( n = 2k \) for some integer \( k \).

2. **Case 1: \( n \) is not a power of 2.**
   - Suppose \( k = 2^a \cdot b \) for an odd integer \( b > 1 \). Then:
     \[
     \frac{2^n - 1}{3} = \frac{2^{2k} - 1}{3} = \frac{(2^k)^2 - 1}{3} = \frac{(4^b)^{2^a} - 1}{3}
     \]
     Using the difference of squares repeatedly, we get:
     \[
     \frac{(4^b)^{2^a} - 1}{3} = \frac{1}{3} \left((4^b)^{2^{a-1}} + 1\right)\left((4^b)^{2^{a-2}} + 1\right) \ldots \left(4^b + 1\right)\left(4^b - 1\right)
     \]
     For \( \frac{2^n - 1}{3} \) to be a divisor of \( 4m^2 + 1 \), we must have \( 2^b - 1 \equiv 3 \pmod{4} \) as a divisor of \( 4m^2 + 1 \). However, this violates Fermat's Christmas Theorem, which states that no number of the form \( 4m^2 + 1 \) can be divisible by 3. Therefore, \( n \) cannot be of this form.

3. **Case 2: \( n \) is a power of 2.**
   - Let \( n = 2^j \) for some integer \( j \). Then:
     \[
     \frac{2^n - 1}{3} = \frac{2^{2^j} - 1}{3}
     \]
     Notice that for \( n = 2^j \), the expression \( \frac{2^n - 1}{3} \) simplifies to a product of Fermat numbers, which are known to be pairwise relatively prime. Hence, we can construct our value of \( m \) using the Chinese Remainder Theorem (CRT) such that:
     \[
     m \equiv 2^{2^c - 1} \pmod{4^{2^c} + 1}
     \]
     for all \( c \in [0, a-1] \). This construction ensures that \( \frac{2^n - 1}{3} \) is a divisor of \( 4m^2 + 1 \).

Thus, the solutions for \( n \) are \( \boxed{2^j \text{ for } j \in \mathbb{N}} \).