Prove the inequality $3^n - 2^{[n \log3/\log2]}>1$ I know one way to find all the integer solutions of $3^n-2^m=1$ is to consider cases about $n$ (odd or even) and proceed consequently. But I would like to prove the above just in a different way:

*

*$m$ cannot be bigger than $n \log_{2}3$ otherwise $3^n-2^m$ would be negative, so $m\le [n \log_{2}3]$


*$m$ cannot be less than $n$ otherwise $3^n-2^m$ would be bigger than $1$


*As between the integers $2^m$ and $2^{m+1}$ should be one prime then we must have $m+1> n \log_{2}3$
From 1., 2. and 3. we get that $m= [n \log_{2}3]$
So in order to prove that the only solutions are $n=1$ and $m=1$ or $n=2$ and $m=3$, it suffices to prove:
$$3^n-2^{[n \log_{2}3]}>1$$ for every integer $n>2$.
But I have difficulty here and I need some advise please.
 A: Note that $3-2=1$ and $3^2-2^3=1$. Those are the only solutions.
ETA: I have no idea how one would prove it along the lines that you are trying to, showing that there are no integers $N$ and $M$ s.t. $0<M-\frac{N \log 2}{\log 3} \le \frac{1}{3^M}$ via some sort of analytic method [i.e., $2^N = 3^{\frac{N \log 2}{\log 3}}$ is an integer, and
$\frac{N \log 2}{\log 3}$ comes so close to another integer $M$ from below that in fact $3^M-3^{\frac{N \log 2}{\log 3}} = 1$]. It probably would involve some pretty deep mathematical results.
Are you aware of the other proof first: First note that the equation $$3^M-2^N=1$$ has no solutions for $M>1$ odd. Indeed if $M$ is at least 3 then $N$ must also satisfy $N>3$. However, $3^M \equiv_8 3$ for all odd $M$ while $2^N \equiv_8 0$ for all $N \ge 3$.
We now show that the equation $$3^{M}-2^N=1$$ has no nontrivial solutions either
for any even integer $M>2$.
Indeed doing algebra this becomes
$$(3^{\frac{M}{2}}-1)(3^{\frac{M}{2}}+1)=2^N.$$
Then this implies that there exists nonnegative integers $n_1$, $n_2$ such that both (a): $3^{\frac{M}{2}}-1=2^{n_1}$ and (b): $3^{\frac{M}{2}}+1 = 2^{n_2}$, and $n_1+n_2=N$, and [as $3^{\frac{M}{2}}-1<3^{\frac{M}{2}}+1$] the strict inequality $n_1<n_2$. So then, on the one hand $2^{\min\{n_1,n_2\}}=2^{n_1}=3^{\frac{M}{2}}-1$ divides both $3^{\frac{M}{2}}-1$ and $3^{\frac{M}{2}}+1$ by Equations (a) and (b), so (c): $2^{n_1} =3^{\frac{M}{2}}-1$ is a factor of $\gcd(3^{\frac{M}{2}}-1,3^{\frac{M}{2}}+1)$. On the other hand, $3^{\frac{M}{2}}-1$ and $3^{\frac{M}{2}}+1$ differ by precisely 2 so this implies (d): $\gcd(3^{\frac{M}{2}}-1,3^{\frac{M}{2}}+1) \in \{1,2\}$.  So (c) and (d) together give $3^{\frac{M}{2}}-1$ must be in $\{1,2\}$ if there is a solution, which is impossible for even $M \geq 4$.
