Integer solutions of $3^n-1=2m^2$

According to Dickson's "History of the Theory of Numbers" vol. 2, pg. 694,

"E. Fauquembergue [158] proved that $1+3+3^2+ · · · +3^n=y^2$ only when n=0, 1, 4, by using the powers of $a+b\sqrt{-2}$ to treat $3^{n+1}=1+2y^2$."

The reference is to "Mathesis, (2), 4, 1894, 169-170".

This is only to say one can disregard the case of $n \ge 4$ even, using Catalan's conjecture/theorem.

For even $n=2k\ge 4,$ if $3^n-1-2m^2$ then $(3^k-1)(3^k+1)=2m^2,$ where the factors on the left have gcd $2.$ This means one factor is $2a^2$ and the other is $b^2,$ where (unneeded) $\gcd(a.b)=1.$ So we get either $3^k-1=b^2$ or $3^k+1=b^2,$ where now $k \ge 2.$

We have then a difference of nontrivial powers equal to $1.$ Now apply Catalan's conjecture (theorem). We'll have either $3^k-b^2=1$ or $b^2-3^k=1,$ and neither fits the only solution (Catalan) $2^3-3^2=1.$