Only cubes and squares!

I'm somewhat questioning the logic in your third line. It turns out to be correct, but I have some doubts about how you were able to prove it. Here is my line of thinking:

By observing the original equation mod $9$ , It is observed that $x^3\equiv 0\pmod{9}$ . However, from what I can tell, this does not prove that $x\equiv 0\pmod{9}$ . It merely proves that $x\equiv 0\pmod{3}$ (because any cube of a multiple of $3$ is divisible by $9$ ).

Let $x=3p$ . The equation becomes $27p^3-36p^3=5184$ . Dividing both sides of the equation by $9$ yields $3p^3-4y^3=576$ . Now observe this equation mod $3$ . This gives $y^3\equiv 0\pmod{3}$ . Therefore $y\equiv 0\pmod{3}$ .

Let $y=3q$ . The equation becomes $3p^3-108q^3=576$ . Dividing both sides of the equation by $3$ yields $p^3-36q^3=192$ .

Using a similar process as before, you can prove that $p\equiv 0\pmod{3}$ . Then let $p=3r$ (This now proves that $x$ is divisible by $9$ ). This gives the equation $27r^3-36q^3=192$ . Dividing both sides of the equation by $3$ yields $9r^3-12q^3=64$ . Finally, it can be observed that the left side of the equation is divisible by $3$ , while the right side is not. Therefore, there are no solutions.