This does not completely answer the question. This provides some more information than those given by previous answers and comments. Also, the method I provide here slightly differs from the following reference, but main ideas are from the paper.
According to the reference http://www.fq.math.ca/Papers1/43-2/paper43-2-6.pdf
We obtain common prime divisors of $2^n-3$ and $3^n-2$ in the following way:
Find a prime number $p$ satisfying
$$
\gcd(\mathrm{ord}_p (3\cdot 2^{-1}), \mathrm{ord}_p (6)) = 1 \ \mathrm{or} \ 2.
$$
For such prime $p$, let $\ell = \mathrm{ord}_p(3\cdot 2^{-1})$ and $k=\mathrm{ord}_p(6)$. Find integer solutions to:
$$
\ell x - k y = 2.$$
We want $x\geq 1$ from Step 2. Starting from $1$, find:
$$
2^{\ell x -1} \ \mathrm{mod} \ p, \ \mathrm{and} \ 3^{\ell x -1} \ \mathrm{mod} \ p
$$
If they are $3$ and $2$ respectively, then the prime $p$ is the common divisor of $2^n-3$ and $3^n-2$ where $n=\ell x -1$.
A SAGE program that I wrote, doesn't exactly follow the above algorithm, but it was successful in finding at least the three primes $5$, $5333$, $18414001$.
for p in primes(4,40000000):
a=2.inverse_mod(p);
b=Mod(Mod(3,p)*a,p).multiplicative_order();
c=Mod(6,p).multiplicative_order();
if gcd(b, c)==1:
g,s,t=xgcd(b, -c);
h=(2*s)%c;
print p, Mod(b*h-1,4), power_mod(3,b*h-1,p), power_mod(2,b*h-1,p);
with the following results:
5 3 2 3
5333 1 5331 5330
18414001 3 2 3
The second line result may be due to the program lacking to proceed Step 2 and Step 3. With more care, it might be possible to implement those and get a correct result 5333 3 2 3.
It was not possible for me to completely settle the (equivalent) problem:
$\bullet$ If $p|\gcd(2^n-3,3^n-2)$, then $n\equiv 3 \ \mathrm{mod} \ 4$.
But, I now have one more conjecture:
$\bullet$ If $\gcd(\mathrm{ord}_p (3\cdot 2^{-1}), \mathrm{ord}_p (6)) = 1$, then there exists $n$ such that $p|\gcd(2^n-3,3^n-2)$.
Both of these questions still remain open.