Rewrite the equation as:
$$3x^4 + 2013 + 24x^2 = 25y^2$$
which gives
$$25y^2 = 3x^4 + 24x^2 + 2013$$
So $25y^2$ is congruent to $3x^4 + 24x^2 + 2013$.

Let us consider the equation modulo $25$:
$$3x^4 + 24x^2 + 2013 \equiv 0 \pmod{25}$$
Note that $2013 \equiv 13 \pmod{25}$, so:
$$3x^4 + 24x^2 + 13 \equiv 0 \pmod{25}$$
Let $x^2 = t$, then $x^4 = t^2$:
$$3t^2 + 24t + 13 \equiv 0 \pmod{25}$$
We can check all possible values of $t$ modulo $25$ (since $x$ is integer, $t$ is a quadratic residue modulo $25$).

The quadratic residues modulo $25$ are $0, 1, 4, 9, 16, 24, 6, 19, 11, 24, 6, 19, 11, 24, 6, 19, 11, 24, 6, 19, 11, 24, 6, 19, 11$ (but let's just check $t = 0$ to $24$).

For $t = 0$:
$$3 \cdot 0^2 + 24 \cdot 0 + 13 = 13 \not\equiv 0 \pmod{25}$$
For $t = 1$:
$$3 \cdot 1 + 24 \cdot 1 + 13 = 3 + 24 + 13 = 40 \equiv 15 \pmod{25}$$
For $t = 4$:
$$3 \cdot 16 + 24 \cdot 4 + 13 = 48 + 96 + 13 = 157 \equiv 7 \pmod{25}$$
For $t = 9$:
$$3 \cdot 81 + 24 \cdot 9 + 13 = 243 + 216 + 13 = 472 \equiv 22 \pmod{25}$$
For $t = 16$:
$$3 \cdot 256 + 24 \cdot 16 + 13 = 768 + 384 + 13 = 1165 \equiv 15 \pmod{25}$$
For $t = 24$:
$$3 \cdot 576 + 24 \cdot 24 + 13 = 1728 + 576 + 13 = 2317 \equiv 17 \pmod{25}$$
None of these are congruent to $0$ modulo $25$.

Therefore, there is no integer $x$ such that $3x^4 + 24x^2 + 2013$ is divisible by $25$, so there are no integer solutions to the equation.