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! $:)$
 A: Here is a very short solution for the second one:
By little Fermat and Euler's totient theorem we know that with some nonnegative integers k and j
$$ 2^{2 \cdot 3^j k} - 1 \equiv 0 \pmod{ 3^{j+1}} $$
Thus
$$(2^{3^j k})^2 - 1 \equiv 0 \pmod{ 3^{j+1}} \\ 
(2^{3^j k} - 1)(2^{3^j k} + 1) \equiv 0 \pmod{ 3^{j+1}} $$
But here the factors on the lhs differ by 2 and thus cannot both be divisible by 3, and so must have other factors besides powers of 3, except for the case where one parenthese equals 1. So we can determine the single solution : $j=0,k=1 \to (1)(3) =3^1=2^2-1 $ [edit: corrected $\to k=1$]
A: 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 $i=3$.
A: 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 $ord_{(3^n)}2=2\cdot3^{n-1}=\phi(3^n)$
]
