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 $n$.
Analysis We first guess the value of $n$ by specific assignments, then proceed to prove it.
Solution When $n=1$, $\left|2^{1}-3^{1}\right|=1 ; \quad n=5$ when, $\left|2^{2}-3^{2}\right|=5 ; \quad n=7$ when, $\left|2^{1}-3^{2}\right|=7$;
$n=11$ when, $\left|2^{4}-3^{3}\right|=11 ; \quad n=13$ when, $\left|2^{4}-3^{1}\right|=13 ; \quad n=17$ when, $\left|2^{6}-3^{4}\right|=17$;
$n=19$ when, $\left|2^{3}-3^{3}\right|=19 ; \quad n=23$ when, $\left|2^{5}-3^{2}\right|=23 ; \quad n=25$ when, $\left|2^{1}-3^{3}\right|=25$;
$n=29$ when, $\left|2^{5}-3^{1}\right|=29 ; \quad n=31$ when, $\left|2^{5}-3^{0}\right|=31$;

Next, we prove that $n=35$ satisfies the requirement using proof by contradiction. 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 $a \neq 0,1,2$, so $a \geqslant 3$. Modulo 8, we get $-3^{b} \equiv 3(\bmod 8)$, i.e., $3^{b} \equiv 5(\bmod 8)$, but $3^{\prime}=1,3(\bmod 8)$, which is impossible.
(2) If $3^{b}-2^{a}=35$, it is easy to see that $b \neq 0,1$. Modulo 9, we get $2^{a} \equiv 1(\bmod 9)$,
since $\left\{2^{k}(\bmod 9)\right\}: 2,4,8,7,5,1,2,4, \cdots$,
thus $2^{6 k} \equiv 1,2^{6 k+1} \equiv 2,2^{6 k+2} \equiv 4,2^{6 k+3} \equiv 8,2^{6 k+4} \equiv 7,2^{6 k+5} \equiv 5(\bmod 9)$.
Therefore, $a=6 k$, where $k$ is a non-negative integer, so $3^{b}-8^{2 k}=35$.
Modulo 7, we get $3^{b} \equiv 1(\bmod 7)$,
since $\left\{3^{k}(\bmod 7)\right\}: 3,2,6,4,5,1,3,2, \cdots$, hence $b=6 k^{\prime}$, where $k^{\prime}$ is a positive integer,
so $3^{6 k^{k}}-2^{6 k}=35,\left(3^{3 k^{k}}-2^{3 k}\right)\left(3^{3 k^{k}}+2^{3 k}\right)=35$,
thus $\left\{\begin{array}{l}3^{3 k^{\prime}}-2^{3 k}=1 \\ 3^{3 k^{\prime}}+2^{3 k}=35\end{array}\right.$ or $\left\{\begin{array}{l}3^{3 k^{k}}-2^{3 k}=5 \\ 3^{3 k^{\prime}}+2^{3 k}=7\end{array}\right.$,
therefore, $3^{3 k^{\prime}}=18$ or 6, which is impossible.
In conclusion, $n_{\min }=35$.