For which integer values of $x$ and $y$ do we have: $7^{x}-3 \times 2^{y}=1$?
Let's first write the equation in the form: $7^{x}-1=3 \times 2^{y}$. The solution involves factorizations, notably: $7^{x}-1=(7-1)\left(7^{x-1}+7^{x-2}+\cdots+7+1\right)$. Since $7-1=3 \times 2$, the equation becomes: $7^{x-1}+7^{x-2}+\cdots+7+1=2^{y-1}$. Either $y=1$, which provides a first solution $(x=y=1)$, or the right-hand side is even and the left-hand side contains an even number of terms (all odd), so $x$ is even. In this case, we can factorize again by $(7+1)=2^{3}$, which brings us to: $7^{x-2}+7^{x-4}+\cdots+7^{2}+1=2^{y-4}$. Once again, either $y=4$ which provides a second solution $(x=2)$, or both sides are even, so the first contains an even number of terms, $x$ is a multiple of 4. We can factorize the first member again as: $\left(7^{2}+1\right)\left(7^{x-4}+7^{x-8}+\ldots+7^{4}+1\right)$. But this time, $7^{2}+1=50$ cannot divide the second member which is a power of 2, so there is no solution for $y>4$. The only solutions are $(1,1)$ and $(2,4)$.