The answer is no. Actually [Catalan's conjecture][1], or Mihăilescu's theorem, suggests that the only solution of your equation $2^k = 3^z - 1$ is $k=3,z=2$.

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As the comment below has mentioned, I missed another solution $k=z=1$.


  [1]: https://en.wikipedia.org/wiki/Catalan%27s_conjecture