Example 7 A positive integer $n$ cannot be divisible by $2$ or $3$, and there do not exist non-negative integers $a$, $b$ such that $\left|2^{a}-3^{b}\right|=n$. Find the minimum value of $n$.
(2003, National Training Team Problem)
Explanation: When $n=1$, $\left|2^{1}-3^{1}\right|=1$;
When $n=5$, $12^{2}-3^{2} \mid=5$;
When $n=7$, $\left|2^{1}-3^{2}\right|=7$;
When $n=11$, $\left|2^{4}-3^{3}\right|=11$;
When $n=13$, $\left|2^{4}-3^{1}\right|=13$;
When $n=17$, $12^{6}-3^{4} \mid=17$;
When $n=19$, $12^{3}-3^{3} \mid=19$;
When $n=23$, $12^{5}-3^{2} \mid=23$;
When $n=25$, $12^{1}-3^{3} \mid=25$;
When $n=29$, $12^{5}-3^{1} \mid=29$;
When $n=31$, $\left|2^{5}-3^{0}\right|=31$.
Below, we use proof by contradiction to show that $n=35$ satisfies the requirement.
If not, there exist non-negative integers $a, b$ such that
$$
\left|2^{a}-3^{b}\right|=35.
$$
(1) If $2^{a}-3^{b}=35$, clearly, $a \neq 0,1,2$, hence $a \geqslant 3$. Taking modulo 8, we get $-3^{b} \equiv 3(\bmod 8)$, i.e., $3^{b} \equiv 5(\bmod 8)$, but $3^{b} \equiv 1,3(\bmod 8)$, which is impossible.
(2) If $3^{b}-2^{a}=35$, it is easy to see that $b \neq 0,1$. Taking modulo 9, we get $2^{a} \equiv 1(\bmod 9)$. And
$$
\begin{array}{l}
2^{6 k} \equiv 1(\bmod 9), 2^{6 k+1} \equiv 2(\bmod 9), \\
2^{6 k+2} \equiv 4(\bmod 9), 2^{6 k+3} \equiv 8(\bmod 9), \\
2^{6 k+4} \equiv 7(\bmod 9), 2^{6 k+5} \equiv 5(\bmod 9).
\end{array}
$$

Thus, $a=6 k(k \in \mathbf{N})$.
Therefore, $3^{b}-8^{2 k}=35$.
Taking modulo 7, we get $3^{b} \equiv 1(\bmod 7)$.
And $\left\{3^{k}(\bmod 7)\right\}: 3,2,6,4,5,1,3,2,6,4$, $5,1, \cdots$, hence $b=6 k^{\prime}\left(k^{\prime} \in \mathbf{N}_{+}\right)$. Therefore, $3^{6 k^{\prime}}-2^{6 k}=35$,

i.e., $\left(3^{3 k^{\prime}}-2^{3 k}\right)\left(3^{3 k^{\prime}}+2^{3 k}\right)=35$.
Thus, $\left\{\begin{array}{l}3^{3 k^{\prime}}-2^{3 k}=1, \text { or }\left\{\begin{array}{l}3^{3 k^{\prime}}-2^{3 k}=5, \\ 3^{3 k^{k}}+2^{3 k}=35\end{array} 2^{3 k^{k}}+2^{3 k}=7 .\right.\end{array}\right.$
Therefore, $3^{3 k^{\prime}}=18$ or 6, which is impossible.
In conclusion, the smallest value of $n$ is 35.