Post History

First I plug in $k = 1,...,4$ and only find $k=1$ as a solution. Hence I conjecture that there are not solutions apart from $k = 1$

Before we prove this we should recall the geometric series formula
$$\sum_{i = 0}^k 9^i = \frac{9^k - 1}{9 - 1}.$$ (prove this via induction or using polynomials)

This means we are solving the equation $$(9^k - 1) = 8n^2,$$ for positive integers $n$ and $k$.

Notice that $9^k = (3^k)^2$, so we can factor the left hand side to get
$$(3^k + 1)(3^k - 1) = 8n^2.$$

We compute $$\gcd(3^k + 1, 3^k - 1) = \gcd(2, 3^k - 1) = 2.$$
This means that one of the two factors is $2a^2$ and the other factor is $4b^2$, where $a$ is odd.

Looking mod $3$ we see that $3^k + 1 \neq 2a^2$, so we need to solve
$$3^k - 1 = 2a^2,$$
$$3^k + 1 = 4b^2.$$

We can rearrange the second equation to get
$$3^k = (2b + 1)(2b - 1).$$
Since $3$ is prime, we see that both factors are powers of $3$. Since they differ by $2$ we must have $2b + 1 = 3$ and $2b - 1 = 1$. This corresponds to the solution $a = b = n = k = 1$ and this is therefore the only solution.