Is there a quick (hopefully elementary) way to prove that $6b^2c^2 + 3c^2 - 36bc - 4b^4 - 4b^2 + 53=0$ has only one solution? I have the Diophantine equation $$6b^2c^2 + 3c^2 - 36bc - 4b^4 - 4b^2 + 53=0.$$ Numerical calculations suggest this has only one positive integer solution, namely $(b,c)=(2,3)$. Is there a quick way to prove or disprove that?

Hint
$$6b^2c^2+3c^2-36bc-4b^4-4b^2+53=(6b^2c^2-36bc+54)-(4b^4+4b^2+1)+3c^2=6(bc-3)^2-(2b^2+1)^2+3c^2=0\\\implies 6(bc-3)^2+3c^2=(2b^2+1)^2$$ Then $3\mid 2b^2+1$. Now proceed like this by checking modulo primes like $2,3$ and so on.
