I was recently considering the puzzle of finding consecutive (integer) powers of 3 that sum to a square. It's not hard to show that this can be reduced to finding values of $n$ such that
$S_3(n) = \displaystyle\sum_{i=0}^n 3^i\tag*{}$
is either a square or one more than a square. It's not hard to find three such values: $S_3(0) = 1$, $S_3(2) = 4$ and $S_3(5) = 121$. However, I've verified numerically that for no other values of $n$ up to $10,000$ is $S_3(n)$ either a square or one more than a square.
How would one go about either proving that these are the only such values for $n$ or finding other values (other than brute force search, assuming there are more values to find)?
The question could obviously be generalized to bases other than 3. When is
$S_b(n) = \displaystyle\sum_{i=0}^nb^i\tag*{}$
either a square or one more than a square. Here are the results of my experimenting for low values of $b$ and $0<n\le10,000$.
$\begin{array}{c|l}b&n\\\hline2\\3&3^0+3^1 = 2^2\\&3^0+3^1+3^2+3^3+3^4 = 11^2\\4&4^0+4^1 = 2^2+1\\5\\6\\7&7^0+7^1+7^2+7^3 = 20^2\\8&8^0+8^1 = 3^2\\9&9^0+9^1 = 3^2+1\end{array}\tag*{}$
We have $$ \sum_{i=0}^n 3^i = \frac{3^{n+1}-1}{2} $$
Therefore, if $\sum_{i=0}^n 3^i=a^2$, then we have $$ 3^{n+1} = 1 + 2a^2 $$
It follows from a result of Nagell that this equation has no solutions for $n\ge 2$ except $a=11$ and $n=4$. ($242+1 = 243$). The remaining cases - $n=0$ and $n=1$ - both give rise to easy solutions.
Note: Nagell's result is actually stronger, and replaces the '$3$' in the above expression with an arbitrary positive integer $y$: i.e., it states that there is no solution to $y^{n+1}=1+2a^2$ for $n\ge 2$ except $y=3,n=4,a=11$. I don't know if there is an easier proof of this special case.
Update: You also asked about the 'one more than a square' case, which we can prove using more elementary means. In this case, we have $\sum_{i=0}^n 3^i = a^2 + 1$ and so
$$ 3^{n+1} = 3 + 2a^2 $$
We know that the number $9$ does not divide the right hand side of this equation: indeed, if $a$ is a multiple of $3$, then $2a^2$ is a multiple of $9$ and therefore $3 + 2a^2$ is not a multiple of nine. Conversely, if $a$ is not a multiple of $3$, then $3+2a^2$ cannot even be a multiple of $3$.
The only possible value of $3^{n+1}$ that is not a multiple of $9$ is $3$ (when $n=0$). From this it follows that $a=0$ too.
Nagell, Trygve, Contributions to the theory of a category of Diophantine equations of the second degree with two unknowns, Nova Acta Soc. Sci. Upsal., Ser. IV 16, No. 2, 38 p. (1954). ZBL0057.28304.