$x^3+48=y^4$ does not have integer (?) solutions How does one find all positive integer solutions to the equation $x^3+48=y^4$?

I realize this is a massive revive, but here is a solution without any powerful theorems about elliptic curves.
We have $x^3 + 64 = y^4 + 2^4$. It is easy to show all (odd) primes dividing the RHS are $1 \pmod{8}$. Now, looking modulo $16$ we deduce either $x \equiv 1 \pmod{16}$ and $y$ is odd or $x,y$ are both even. For the second case if $v_2(x) = 1$ we derive a contradiction because $v_2(LHS) = 3$ while $v_2(RHS) \ge 4$. If $v_2(x) \ge 2$ we have $v_2(LHS) \ge 6$. Thus we must have $v_2(y) = 1$ for $v_2(y^4 + 16) > 4$. Writing $y = 2k$ we need $v_2(k^4 + 1) \ge 2$. But this is absurd since $v_2(k^4 + 1) \le 1$ by checking modulo $4$. Thus it follows $x \equiv 1 \pmod{16}$.
But now utilize $x^3 + 64 = (x+4)(x^2 - 4x + 16)$. $x+4$ is $5 \pmod{8}$, which is absurd so we are done.
