Find all integers $a$ and $b$ such that $7^{a}-3 \times 2^{b}=1$.
We see that $a=0$ or $b=0$ are impossible. We rewrite the equation as follows:

$$
7^{a}-1=6 \times \sum_{i=0}^{a-1} 7^{i}=6 \times 2^{b-1}
$$

If $a=1$, then $b=1$ is the only solution. If $a=2$ then $b=4$ is the only solution.

Suppose now that $a>2$ which implies that $b>4$. Since the sum on the left is even, but consists of odd terms, there must be an even number of terms, so $a$ is even. We group even and odd terms to get

$$
(7+1) \sum_{i=0}^{a / 2-1} 7^{2 i}=8 \times 2^{b-4}
$$

We simplify, and recall that since $b>4$, the sum on the left is even but still consists of odd terms. Therefore, there must be an even number of terms ($a / 2$ is even), and we can repeat the procedure:

$$
\left(7^{2}+1\right) \sum_{i=0}^{a / 4-1} 7^{4 i}=2^{b-4}
$$

This leads to a contradiction because a power of 2 is not divisible by 50.

The only solutions are therefore $(1,1)$ and $(2,4)$.