Proving this $\gcd(2^n-1,3^n+2)=1$ for all postive integers $n$ I have found 
$$\gcd(2-1,3+2)=\gcd(1,5)=1$$
$$\gcd(2^2-1,3^2+2)=\gcd(3,11)=1$$
$$\gcd(2^3-1,3^3+2)=\gcd(7,29)=1$$
$$\gcd(2^4-1,3^4+2)=\gcd(15,83)=1$$
$$\gcd(2^5-1,3^5+2)=\gcd(31,245)=1$$
$$\cdots\cdots$$
I  conjecture $\gcd(2^n-1,3^n+2)=1$,I can't prove this?

I have checked with Sage that, for $n=176$, we have
\begin{align*}
2^n-1 &= 95780971304118053647396689196894323976171195136475135 \\
3^n+2 &= 940461086986004843694934910131056317906479029659199959555574885740211572136210345923,
\end{align*}
and $gcd(2^n-1,3^n+2) = 257$.
$n=176$ is the smallest. The next ones are: 432, 688, 944.
