Are there positive integers $a, b$ with $b \ge 2$ such that $2^a + 1$ is divisible by $2^b - 1$?
1. We start by considering the problem statement: Are there positive integers \(a, b\) with \(b \ge 2\) such that \(2^a + 1\) is divisible by \(2^b - 1\)?

2. First, we check the case \(b = 2\). If \(b = 2\), then \(2^b - 1 = 2^2 - 1 = 3\). We need \(2^a + 1\) to be divisible by 3. Notice that \(2^a \equiv -1 \pmod{3}\) if \(a\) is odd. Therefore, for \(b = 2\), any odd \(a\) will work. 

3. Now, let's assume \(b > 2\). The order of 2 modulo \(2^b - 1\) is \(b\). This means that the smallest positive integer \(d\) such that \(2^d \equiv 1 \pmod{2^b - 1}\) is \(d = b\).

4. If \(2^b - 1\) divides \(2^a + 1\), then \(2^a \equiv -1 \pmod{2^b - 1}\). Squaring both sides, we get:
   \[
   2^{2a} \equiv 1 \pmod{2^b - 1}
   \]
   This implies that the order of 2 modulo \(2^b - 1\) divides \(2a\). Since the order is \(b\), \(b\) must divide \(2a\).

5. Since \(b\) divides \(2a\) but not \(a\), \(b\) must be even. Let \(b = 2k\). Then \(2k\) divides \(2a\), which implies \(k\) divides \(a\). Let \(a = mk\) where \(m\) is an odd integer.

6. We need \(2^a \equiv -1 \pmod{2^b - 1}\). Substituting \(a = mk\), we get:
   \[
   2^{mk} \equiv -1 \pmod{2^{2k} - 1}
   \]
   Since \(m\) is odd, we can simplify this to:
   \[
   2^k \equiv -1 \pmod{2^{2k} - 1}
   \]
   This implies:
   \[
   2^k + 1 \geq 2^{2k} - 1
   \]
   Simplifying, we get:
   \[
   2^k + 1 \geq 2^{2k} - 1 \implies 2^k + 2 \geq 2^{2k}
   \]
   For \(k \geq 2\), \(2^k + 2 < 2^{2k}\). Therefore, the only possible value for \(k\) is 1.

7. If \(k = 1\), then \(b = 2k = 2\). This is the only possibility that satisfies the conditions.

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