Find all positive integers $m, n$ such that $7^{m}-3 \cdot 2^{n}=1$.
We look at the cases $n=0,1$ which give the solution (1,1). Then modulo 4 we find that $m$ is even, so we can write

$$
7^{2 m^{\prime}}-1=\left(7^{m^{\prime}}-1\right)\left(7^{m^{\prime}}+1\right)=3 \cdot 2^{n}
$$

We note that $m^{\prime} \geqslant 1$ and observe that $6 \mid 7^{m^{\prime}}-1$ by factorization. Thus, with $a \geqslant 1$, we have $\left\{7^{m^{\prime}}-1=3 \cdot 2^{a}=2^{a+2}-2^{a}\right.$ $\left\{7^{m^{\prime}}+1=2^{b}=7^{m^{\prime}}-1+2=2^{a+2}-2^{a}+2\right.$ From this, we deduce that if $a \geqslant 2$ we have a contradiction because the right-hand side of the second equation will be twice an odd number. Therefore, $a=1$ and we obtain the solution $(2,4)$.

Another solution using Zsigmondy: we rewrite this as $7^{m}-1=3 \cdot 2^{n}$, and use Zsigmondy. However, be careful, as the exceptional case is involved! So for $m \geqslant 3$, $7^{m}-1$ has a prime factor that does not divide $7-1=6$, so we restrict ourselves to the case $n=1$, which gives us the solution $(1,1)$, and the case $n=2$ which gives us $(2,4)$.