The equation
$$
\frac{a^m-1}{a-1}=2b^2
$$
does not have solutions in positive integers for $m>2$ as shown below.
First, notice that $a$ must be odd.
Second, one can see that $m$ must be even. Indeed, if $a\equiv 3\pmod4$, then odd $m$ would produce an odd number in the left-hand side. On the other hand, if $a\equiv 1\pmod4$, then from Theorem 3 of the LTE it follows that
$$
\nu_2(m)=\nu_2(\frac{a^m-1}{a-1})=\nu_2(2b^2)>0,
$$
i.e. $m$ is even.
Let $m=2k$ where $k>1$. Then
$$
\frac{a^m-1}{a-1}=\frac{a^k-1}{a-1}(a^k+1).
$$
Since $\gcd(a^k-1,a^k+1)=2$, we have two cases to consider: $\frac{a^k-1}{a-1}=c^2$ or $a^k+1=c^2$ for some $c\mid b$.
In the former case:
if $k=2$, we get $(a,c)=(t^2-1,t)$ for any $t$ satisfying $(t^2-1)^2+1=2d^2$ where $d=b/c$. However, it can be computationally verified (e.g., in Magma) that this elliptic curve has the only solution with $t=0$ (giving $a=-1$).
if $k>2$, we get Ljunggren equation giving $(k,a,c)=(5,3,11)$ and $(k,a,c)=(4,7,20)$, which however do not produce any solutions to the original equation.
In the latter case, one can refer to Mihailescu's Theorem to obtain the only solution $(k,a,c)=(3,2,3)$, which again does not produce any solutions to the original equation.
Hence, the equation in question does not have any solutions for $m>2$.
