Integer solutions of $3^n-1=2m^2$ I want to find solutions of $3^n-1=2m^2$ other than $(n, m)=(0, 0)$, $(n, m)=(1, \pm1)$, $(n, m)=(2, \pm2)$ and $(n, m)=(5, \pm11)$.
There are no other "small" solutions ($n<1000$). For even $n$, we can let $3^{n/2}=k$ and solve Pell equation $k^2-2m^2=1$, and similarly for odd $n$, let $3^{(n-1)/2}=k$ and the equation becomes Pell-like equation $3k^2-2m^2=1$. The complete set of solution is known in both cases, but I cannot prove none of large $k$ can be power of $3$.

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) $2^3-3^2=1.$
