How many solution does the $\displaystyle x^2+xy+y^2=27$ equation have in the set of $\displaystyle \mathbb{Q}$?
There are infinitely many rational solutions to this equation. One family of solutions comes from taking y to have numerator –3, and using the continued fraction expansion of $\displaystyle \sqrt3$. That leads to the solutions $\displaystyle (x,y) = (6,-3),\; \bigl(\frac{69}{13},-\frac3{13}\bigr),\; \bigl(\frac{942}{181},-\frac3{181}\bigr),\; \bigl(\frac{13101}{2521},-\frac3{2521}\bigr),\ldots\,.$
If $\displaystyle 3(4b^2-1) = a^2$, suppose for convenience that $\displaystyle a$ is a multiple of 3, $\displaystyle a = 3c.$ Then $\displaystyle 4b^2-1 = 3c^2$, or $\displaystyle \bigl(\frac{2b}c\bigr)^2 - \frac1{c^2} = 3.$ If $\displaystyle c$ is large, then the left side will be close to $\displaystyle \bigl(\frac {2b}c\bigr)^2$ and so $\displaystyle \frac {2b}c$ must be close to $\displaystyle \sqrt3$. So go to the continued fraction calculator and plug in "3" in the square root box. You will then see a list of the convergents for the continued fraction expansion of $\displaystyle \sqrt3$, and you will notice that every fourth item in the list has an even numerator: $\displaystyle \frac21,\;\frac{26}{15},\;\frac{362}{209},\,\ldots \,.$
Let $\displaystyle 2p_n/q_n$ be the n'th term in that sequence. Then $\displaystyle p_1=q_1=1$ and the sequence grows by the inductive rules $\displaystyle p_{n+1} = 7p_n + 6q_n$, $\displaystyle q_{n+1} = 8p_n + 7q_n$. You should then be able to show by induction that $\displaystyle 4p_n^2-1 = 3q_n^2$. That gives an infinite family of integer solutions to the equation $\displaystyle 4x^2-1 = 3y^2$. Going back to the original problem, $\displaystyle 3(4x^2-1) = (3y)^2$, so that gives infinitely many integers x such that $\displaystyle 3(4x^2-1)$ is a square.