Squares of the form $2^j\cdot 3^k+1$ Squares of the form $2^j\cdot 3^k+1$, for j,k nonnegative.
Is it known if there are infinitely many? And if  $2^j\cdot 3^k+1=N$ is a square, then it must be necessarly a semi-prime?
Do you think that 3-smooth neighbour squares is a good name for these squares?

We may find them all by elementary methods. Assume that $2^j3^k=(n-1)(n+1)$. Since $\gcd(n-1,n+1)\leqslant 2$, we get either $j=0$, $n-1=1$, $n+1=3$, or $j\geqslant 1$, $(n-1)/2$ and $(n+1)/2$ are consecutive 3-smooth numbers. They must be a power of 2 and a power of 3, so we should solve the equations $2^a=3^b\pm 1$. This is not hard.

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*$2^a=3^b+1$. Either $b=0,a=1$ or $b\geqslant 1$, 3 divides $2^a-1$, so $a$ is even and $3^b=2^a-1=(2^{a/2}-1)(2^{a/2}+1)$, the multiples are powers of 3 which differ by 2, so $a=2$, $b=1$.


*$2^a=3^b-1$. Either $a=1$, $b=1$ or $a\geqslant 2$, 4 divides $3^b-1$, $b$ is even, $2^a=(3^{b/2}-1)(3^{b/2}+1)$, the multiples are powers of 2 which differ by 2, $b=2$, $a=3$.
