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The answer is no. Actually Catalan's conjecture, or Mihăilescu's theorem, suggests that the only solution of your equation $2^k = 3^z - 1$ is $k=3,z=2$.
As the comment below has mentioned, I missed another solution $k=z=1$.
The answer is no. Actually Catalan's conjecture, or Mihăilescu's theorem, suggests that the only solution of your equation $2^k = 3^z - 1$ is $k=3,z=2$.
The answer is no. Actually Catalan's conjecture, or Mihăilescu's theorem, suggests that the only solution of your equation $2^k = 3^z - 1$ is $k=3,z=2$.
As the comment below has mentioned, I missed another solution $k=z=1$.
The answer is no. Actually Catalan's conjecture, or Mihăilescu's theorem, suggests that the only solution of your equation $2^k = 3^z - 1$ is $k=3,z=2$.
