Find all $a, b, c \in \mathbb{N}$ such that $2^{a} \cdot 3^{b}=7^{c}-1$.
According to Zsigmondy's theorem, since $7^{1}-1=2 \cdot 3$, for $c \geqslant 3$ (note, $c=2$ is an exception to Zsigmondy because $\left.7+1=8=2^{3}\right), 7^{c}-1$ has a prime divisor different from 2 and 3. Therefore, necessarily, $c \leqslant 2$ and the solutions are $(1,1,1)$ and $(4,1,2)$.