$3^p-2^p$ squarefree? Original question:
$3^n-2^{n-1}$ seems to be squarefree. Is it? 
Answer: No, but amongst the primes dividing one of these numbers, $23$ seems to be a special case: no $23^2$ divide any of them
Are there other case where $p$ divide a $3^n-2^{n-1}$ but $p^2$ does not?
Edited question, based on answers:
It is conjectured that $2^p-1$ is squarefree. 
Could it be that $3^p-1$ is also squarefree for $p\neq2$ and $5$ (where $11^2$ appears)? 
Could it be that $3^p-2^p$ is also squarefree for $p\neq11$ (where $23^2$ appears)? 
Thanks
edit: just saw this Is $(3^p-1)/2$ always squarefree?.
Note: $n$ is a positive natural natural in the original question and $p$ is a prime in my edit.
 A: We can show that there are multiples of $49$ using elementary modular arithmetic techniques.
Let $p$ be a prime greater than $3$ (why?), and seek values of $n$ for which
$$3^n\equiv 2^{n-1}\bmod p^2$$
Multiply by $(3^{-1})^{n-1}$ getting
$$3\equiv (2×3^{-1})^{n-1}\bmod p^2$$
Let us first try $p=5$.  We die because the left side is a nonquadratic residue $\bmod p=5$ and the right side, with $2×3^{-1}\equiv 4\bmod 5$, is a quadratic residue.
Fortunately, for $p=7$ we avoid this contradiction because $2×3^{-1}$ is nonquadratic $\bmod 7$ thus also nonquadratic $\bmod 49$.  We then have
$$3\equiv 17^{n-1}\bmod 49$$
where $17\equiv 3\bmod 7$ is a primitive root in the group of units $\bmod 49$ (the only nonprimitive root $\bmod 49$ congruent to $3\bmod 7$ is $31$), therefore this equation must have positive whole number solutions for $n$.
Will Jagy has identified the minimal solution as $n=38$, so let us check this case $\bmod 49$.  Since units give $1$ when raised to the power of $42$, we may render
$$3^{38}\equiv (3^{-1})^4\equiv 33^4\equiv 11^2\equiv 121\equiv\color{blue}{23\bmod 49}$$
And
$$2^{37}\equiv (2^{-1})^5\equiv 25^5\equiv 25×(-12)^2\equiv 3600\equiv \color{blue}{23\bmod 49}$$
A: Well, I posted too quickly.....for $n=67$ it seems we have 2 factor $11$
