Odd Divisibility

Case(1) If $n \leq 1$

$n=0$ will not satisfy so it is not our solution.

$n=1$ is satisfied $1 |3^1-2^1=1$ .So. $n=1$ is our first solution.

Case(2): if $n >1$

$3^n-2^n \\ =(1+2)^n-2^n\\ =1 +{n \choose 1}2+{n \choose 2} 2^2 + {n \choose 3}2^3 + \cdots + {n \choose n}2^n -2^n \\ =1 +{n \choose 1}2+{n \choose 2} 2^2 + {n \choose 3}2^3 + \cdots +2^n-2^n \\=1 +{n \choose 1}2+{n \choose 2} 2^2 + {n \choose 3}2^3 + \cdots +{n \choose {n-1}}2^{n-1}$

As $3^n-2^n$ is an odd integer then it should not be divisible by even so, $n$ will be odd.

If $n$ is a odd prime then each ${n \choose r}$ terms will contain $n$ because it will not cancel out as it is prime.The expression will become $nk+1$ which is never divisible by $n$

If $n$ is an odd composite number.First of all consider $t_r={n \choose r} 2^r =\dfrac{n!}{r!(n-r)!} 2^r \\ =\dfrac{n(n-1)(n-2)...(n-r+1)}{r!} 2^r \\ =n \dfrac{(n-1)(n-2)...(n-r+1)}{1.2.3.4....(r-1).r} 2^r$ .If $r >n/2$ then we will cancel out $r!$ and will keep $(n-r)!$ else we will keep $r!$ .In the denominator there are $r$ consecutive terms and in the numerator there are also $r$ consecutive terms.If we separate $n$ and keep $(r-1)$ terms.In the denominator we can keep such $(r-1)$ which will cancels out by the $(r-1))$ terms of numerator.So, one will term will left .If the term will prime then there will be at least one factor in the numerator because we select $r \leq n/2$ .If it is odd composite then there will be enough factors are there in the numerator.So , each ${n \choose r} 2^r$ will be divisible by $n$ .

So, again the expression will become $nk+1$ which will never divisible by $n.$

Therefore , only solution is $n=1$

Note This method is not perfect but I want to show how I approached.