That can't be the only condition. Of course $ \frac{3^2-1}{3-1}$ is a square.

Thanks a lot scorpius119 I missed to write that $ b$ is even number.I have edited my post! I am so sorry about this mistake!

We have \[ \frac{b^{n} - 1}{b - 1} = b^{n - 1} + \ldots + b + 1 = k^{2} \textrm{, \ \ } n, k \in \mathbb{N} \textrm{, \ } n \neq 1 \textrm{,} \] from which we can get \[ k^{2} - 1 = b \sum_{i = 0}^{n - 2}{b^{i}} \textrm{.} \] Because $ \displaystyle b \equiv 0 \pmod{2}$, from the left side follows $ \displaystyle k \equiv 1 \pmod{2}$, so $ k = 2t - 1$, $ t \in \mathbb{N}$, $ t \neq 1$ (because $ b \geqslant 2$). Now we get \[ 4t \left( t - 1 \right) = b \sum_{i = 0}^{n - 2}{b^{i}} \textrm{.} \] Notice that $ 4t \left( t - 1 \right) \equiv 0 \pmod{8}$ and \[ \sum_{i = 0}^{n - 2}{b^{i}} \equiv 1 \pmod{2} \textrm{,} \] from which directly follows $ b \equiv 0 \pmod{8}$. Q. E. D.

Easily problem but it is beautiful problem

Lol easy