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! $:)$

Observe that $2$ is a primitive root  of $3^k$ where $k$ is natural number(Proof below).
(1)So if $2^i=3^{j+1}+1,2^i\equiv1\pmod {3^{j+1}}$
$\implies \phi(3^{j+1})\mid i \implies 2\cdot 3^j\mid i$
So, the minimum possible integral value of $i$ is $2\cdot 3^j$
If $j=0,i_{min}=2,2^i-3^{j+1}=2^2-3^1=1$
If $j=1,i_{min}=2\cdot3^1=6,2^i-3^{j+1}=2^6-3^2>1$
If $j=2,i_{min}=2\cdot3^2=18,2^i-3^{j+1}=2^{18}-3^3>1$
Clearly, the difference diverges from $1$ with the increment of $j$ and $i$.
It can also proved as follows: 
$2^i-3^{j+1}-1=2^{2\cdot 3^j}-3^{j+1}-1=4^{3^j}-3^{j+1}$
$=(3+1)^{3^j}-3^{j+1}-1=\sum_{1\le k\le 3^j-1}\binom{3^j}k3^j>0$ if $j>0$
(2) So if $2^i=3^{j+1}-1,2^i\equiv-1\pmod {3^{j+1}}$
$\implies \frac{\phi(3^{j+1})}2\mid i$ but $\phi(3^{j+1})\not\mid i$
So, the minimum possible integral value of $i$ is $3^j$
If $j=0,i_{min}=1,2^i-3^{j+1}=2^1-3^1=-1$
If $j=1,i_{min}=3,2^i-3^{j+1}=2^3-3^2=-1$
If $j=2,i_{min}=3^2=9,2^i-3^{j+1}=2^9-3^{2+1}>-1$
Clearly, the difference diverges from $-1$ with the increment of $j$ and $i$.
[
Proof:
$2\equiv -1\pmod 3, 2^2\equiv1\implies ord_32=2$
$2^2=4,2^3=8\equiv-1\pmod {3^2},2^6\equiv1\implies ord_{(3^2)}2=6=2\cdot 3=\phi(9)$
From this, $ord_{(3^3)}2=2\cdot 3^2=\phi(27)$
As $ord_{(3^2)}2=6=2\cdot 3,ord_{(3^3)}2=2\cdot 3^2\implies ord_{(3^4)}2=2\cdot3^3=\phi(81)$
Proceeding this way, we can prove $ord_{(3^n)}2=2\cdot3^{n-1}=\phi(3^n)$
]
