The value of
$$\left(1-\frac{1}{2^2-1}\right)\left(1-\frac{1}{2^3-1}\right)\left(1-\frac{1}{2^4-1}\right)\dots\left(1-\frac{1}{2^{29}-1}\right)$$
can be written as $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $2m - n.$
1. We start with the given expression:
   \[
   \left(1-\frac{1}{2^2-1}\right)\left(1-\frac{1}{2^3-1}\right)\left(1-\frac{1}{2^4-1}\right)\dots\left(1-\frac{1}{2^{29}-1}\right)
   \]

2. Simplify each term inside the product:
   \[
   1 - \frac{1}{2^k - 1}
   \]
   for \( k = 2, 3, 4, \ldots, 29 \).

3. Notice that \( 2^k - 1 \) can be factored as:
   \[
   1 - \frac{1}{2^k - 1} = \frac{2^k - 2}{2^k - 1} = \frac{2(2^{k-1} - 1)}{2^k - 1}
   \]

4. Rewrite the product using the simplified terms:
   \[
   \prod_{k=2}^{29} \left( \frac{2(2^{k-1} - 1)}{2^k - 1} \right)
   \]

5. Factor out the constant 2 from each term:
   \[
   2^{28} \prod_{k=2}^{29} \left( \frac{2^{k-1} - 1}{2^k - 1} \right)
   \]

6. Observe that the product telescopes:
   \[
   \prod_{k=2}^{29} \left( \frac{2^{k-1} - 1}{2^k - 1} \right) = \frac{1}{2^2 - 1} \cdot \frac{2^2 - 1}{2^3 - 1} \cdot \frac{2^3 - 1}{2^4 - 1} \cdots \frac{2^{28} - 1}{2^{29} - 1}
   \]

7. Simplify the telescoping product:
   \[
   \frac{1}{3} \cdot \frac{3}{7} \cdot \frac{7}{15} \cdots \frac{2^{28} - 1}{2^{29} - 1} = \frac{1}{2^{29} - 1}
   \]

8. Combine the results:
   \[
   2^{28} \cdot \frac{1}{2^{29} - 1} = \frac{2^{28}}{2^{29} - 1}
   \]

9. Let \( m = 2^{28} \) and \( n = 2^{29} - 1 \). Since \( m \) and \( n \) are relatively prime, we need to find \( 2m - n \):
   \[
   2m - n = 2 \cdot 2^{28} - (2^{29} - 1) = 2^{29} - 2^{29} + 1 = 1
   \]

The final answer is \( \boxed{1} \).