To solve the equation $\log_2 (3z - 14) = 1 + \log_2 (z + 1)$, follow these steps:

1. Use the properties of logarithms: $\log_2 (3z - 14) = \log_2 (2) + \log_2 (z + 1)$

2. Combine the logarithms on the right: $\log_2 (3z - 14) = \log_2 (2(z + 1))$

3. Since the logarithms with the same base are equal, set the arguments equal: $3z - 14 = 2(z + 1)$

4. Solve for $z$: $3z - 14 = 2z + 2$ $3z - 2z = 2 + 14$ $z = 16$

5. Check the solution to ensure it lies within the domain of the original logarithmic expressions:

   • $3z - 14 = 3(16) - 14 = 48 - 14 = 34$, which is positive.
   • $z + 1 = 16 + 1 = 17$, which is positive.

Both expressions are positive, so $z = 16$ is valid.

Thus, the exact solution set is $\{16\}$.