Solutions for $ 2^i = 3^ j - 1 $ or $ 2^i = 3^ j + 1 $ Are there any solutions for $ 2^i = 3^ j - 1 $ or $ 2^i = 3^ j + 1 $,
for $i>3$ and $j>2$ ?
Thanks! $:)$

Gottfried Helms has dealt with $2^i=3^j+1$. For completeness we deal with $2^i =3^j-1$.
This has the solutions $i=j=1$ and $i=3$, $j=2$. We show there are no others. 
First we deal with odd $j$. Note that
$$3^j-1=(3-1)\left(3^{j-1}+3^{j-2}+\cdots +1\right).$$
The term $3^{j-1}+3^{j-2}+\cdots +1$ is the sum of an odd number of odd numbers, so it is odd. It follows that if $j$ is odd, then $3^j-1$ can only be a power of $2$ if $j=1$.
Next we deal with even $j$, say $j=2k$. Then $3^j=3^{2k}-1=(3^k-1)(3^k+1)$. This can be a power of $2$ only if both $3^k-1$ and $3^k+1$ are powers of $2$. But the only two powers of $2$ that differ by $2$ are $2$ and $4$. It follows that $j=2$ and $i=3$.
