$x^3+48=y^4$ does not have integer (?) solutions

One way would be to note that a solution of this equation would also give a solution of the equation $x^3 + 48 = y^2$. That's an elliptic curve of rank 1 and with trivial torsion subgroup. There are bounds on the heights of the integral points on a Weierstrass model of an elliptic curve in terms of the coefficients. You can find a generator for the Mordell-Weil group and using the height bounds find all the integral points. In this case, the only point will be $(x,y) = (1,7)$. Since 7 is not a square, you deduce that the original equation has no integral solutions. Of course, I didn't actually compute the bounds, but asked Magma, which does it automatically. Here is the code:

> P<x>:=PolynomialRing(Integers());
> E:=EllipticCurve(x^3 + 48);
> Rank(E);
1
> TorsionSubgroup(E);
Abelian Group of order 1
> IntegralPoints(E);
[ (1 : 7 : 1) ]
[ <(1 : 7 : 1), 1> ]

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.