Let $m$ and $n$ be relatively prime positive integers. Determine all possible values of \[\gcd(2^m - 2^n, 2^{m^2+mn+n^2}- 1).\]
1. Let \( m \) and \( n \) be relatively prime positive integers. We need to determine the possible values of \(\gcd(2^m - 2^n, 2^{m^2+mn+n^2} - 1)\).

2. Without loss of generality, assume \( m \geq n \). We can use the property of gcd for powers of 2:
   \[
   \gcd(2^a - 1, 2^b - 1) = 2^{\gcd(a, b)} - 1
   \]
   Applying this property, we get:
   \[
   \gcd(2^m - 2^n, 2^{m^2+mn+n^2} - 1) = \gcd(2^{m-n} - 1, 2^{m^2+mn+n^2} - 1)
   \]

3. Next, we need to find \(\gcd(m-n, m^2 + mn + n^2)\). Since \(m\) and \(n\) are relatively prime, we can use the Euclidean algorithm:
   \[
   \gcd(m-n, m^2 + mn + n^2)
   \]
   Notice that:
   \[
   m^2 + mn + n^2 = (m-n)(m+n) + mn + n^2
   \]
   Therefore:
   \[
   \gcd(m-n, m^2 + mn + n^2) = \gcd(m-n, 3mn)
   \]

4. Since \(m\) and \(n\) are relatively prime, \(\gcd(m, n) = 1\). Thus, \(\gcd(m-n, 3mn)\) must divide 3. The possible values for \(\gcd(m-n, 3mn)\) are 1 and 3.

5. If \(\gcd(m-n, 3mn) = 1\), then:
   \[
   \gcd(2^{m-n} - 1, 2^{m^2+mn+n^2} - 1) = 2^1 - 1 = 1
   \]

6. If \(\gcd(m-n, 3mn) = 3\), then:
   \[
   \gcd(2^{m-n} - 1, 2^{m^2+mn+n^2} - 1) = 2^3 - 1 = 7
   \]

Therefore, the only possible values for \(\gcd(2^m - 2^n, 2^{m^2+mn+n^2} - 1)\) are 1 and 7.

The final answer is \(\boxed{1 \text{ and } 7}\).