$a^b+2$ or $a^b-2$ is in set
The answer is yes.
Take four odd numbers $a<b<c<d$ from the set, all greater than $3$, such that $c=b^a\pm2$ and $d=c^b\pm2$.
Case 1: $c=b^a\pm2$ and $d=c^b\pm2$ (with equal signs).
Then $c=b^a\pm2\equiv b\pm2 \pmod3$ and $d=c^b\pm2\equiv c\pm2 \pmod3$. One of $b,c,d$ is divisible by $3$.
Case 2: $c=b^a\pm2$ and $d=c^b\mp2$ (with opposite signs). If $b$ is a prime then, by Fermat's theorem, $$ d = c^b \mp2 \equiv c\mp2 = b^a \equiv 0 \pmod{b} $$ so $d$ is divisible by $b$. (Obviously $d$ is greater than $b$.)