For which integer $x,y$ does $x^2+23=3(2^y)$ How do I solve this Lesbegue-Ramanujan-Nagell type equation ($x^2+D=AB^y$):
$x^2+23=3(2^y)$

I have been trying for quite some time now, to no end. Any suggestions/help would be greatly appreaciated.
Clearly $y\ge 0$. Three cases:

$1)$ $y=3k$. Then $(3x)^2+207=\left(3\cdot 2^k\right)^3$. But $a^2+207=b^3$ has $14$ integral solutions (see http://oeis.org/A081120, in particular http://oeis.org/A081120/b081120.txt), which can be found with a program or in this table:

$$(a,b)=(\pm3,6), (\pm39,12), (\pm75,18), (\pm172,31),$$

$$(\pm5511,312), (\pm6022,331), (\pm223063347, 367806)$$

This gives the solutions $(x,y)=(\pm1,3),(\pm13,6)$.

$2)$ $y=3k+1$. Then $(6x)^2+828=\left(3\cdot 2^{k+1}\right)^3$. But $a^2+828=b^3$ has $10$ integral solutions (http://oeis.org/A081120/b081120.txt), which with a program or this table are:

$$(a,b)=(\pm30,12), (\pm37, 13), (\pm114,24), (\pm1122,108), (\pm298254, 4464)$$

This gives the solutions $(x,y)=(\pm 5,4),(\pm 19,7)$.

$3)$ $y=3k+2$. Then $x^2\equiv 3\pmod{7}$, contradiction, because $3$ is not a quadratic residue mod $7$.

Remark: this uses the results from $1998$ (from this paper), namely the full solutions of Mordell's equation $x^2+k=y^3$ for $0&lt;|k|\le 10^4$.
Ill give a Hint for a similar example . See Ramanujan-Nagell equation on the net it is clearly mentioned there that the numbers cannot be $n= 3,4,5,7, 15$ and some more . to $2^n-7=x^2$ the solutions to x  correspond to $1,3,5,11, 181$ and there exists no solution greter than $2^{15}$. for more info you can look up for http://mathworld.wolfram.com/RamanujansSquareEquation.html Hope you get your hints and solution from there.