The answer is $41$. We have:
$2=3^1-2^0$;
$3=2^2-3^0$;
$5=2^3-3^1$;
$7=2^3-3^0$;
$11=3^3-2^4$;
$13=2^4-3^1$;
$17=3^4-2^6$;
$19=3^3-2^3$;
$23=3^3-2^2$;
$29=2^5-3^1$;
$31=2^5-3^0$;
$37=2^6-3^3$.
To prove that $41$ can't be represented in the form $\pm (2^x-3^y)$ we have to consider the two cases:
If $41=2^x-3^y\Rightarrow x\ge 3\Rightarrow 3^y=2^x-41\equiv -1\ (mod\ 8)$ but $3^y\equiv 1,3\ (mod\ 8)$, a contradiction.
If $41=3^y-2^x\Rightarrow x\ge 2$.
$2^x=3^y-41\equiv 1\ (mod\ 3)\Rightarrow x=2u$.
$3^y=2^x+41\equiv 1\ (mod\ 4)\Rightarrow y=2v$.
So, $41=3^{2v}-2^{2u}=(3^v-2^u)(3^v+2^u)\Rightarrow 3^v-2^u=1,\ 3^v+2^u=41$, a contradiction.
Hense, the answer is $41$.