Let us rewrite the equation:
$$x^4 + 68 = 4y^4$$
$$x^4 - 4y^4 = -68$$
$$(x^2 - 2y^2)(x^2 + 2y^2) = -68$$

Now, $-68$ factors as $(-1) \times 68$, $(-2) \times 34$, $(-4) \times 17$, $(-17) \times 4$, $(-34) \times 2$, $(-68) \times 1$ and their negatives. We consider all pairs $(a, b)$ such that $a \times b = -68$.

Let $x^2 - 2y^2 = a$, $x^2 + 2y^2 = b$, so $a + b = 2x^2$, $b - a = 4y^2$.

Thus,
$$x^2 = \frac{a + b}{2}$$
$$y^2 = \frac{b - a}{4}$$

We need $x^2$ and $y^2$ to be integers, so $a + b$ is even and $b - a$ is divisible by $4$.

Let us check all possible pairs $(a, b)$:

1. $a = -1$, $b = 68$
- $a + b = 67$ (odd), $b - a = 69$ (not divisible by $4$)
2. $a = -2$, $b = 34$
- $a + b = 32$ (even), $b - a = 36$ (divisible by $4$)
- $x^2 = 16$, $y^2 = 9$
- $x = \pm 4$, $y = \pm 3$
3. $a = -4$, $b = 17$
- $a + b = 13$ (odd), $b - a = 21$ (not divisible by $4$)
4. $a = -17$, $b = 4$
- $a + b = -13$ (odd), $b - a = 21$ (not divisible by $4$)
5. $a = -34$, $b = 2$
- $a + b = -32$ (even), $b - a = 36$ (divisible by $4$)
- $x^2 = -16$ (not a square), $y^2 = 9$
6. $a = -68$, $b = 1$
- $a + b = -67$ (odd), $b - a = 69$ (not divisible by $4$)

Now, the negatives:
7. $a = 1$, $b = -68$
- $a + b = -67$ (odd), $b - a = -69$ (not divisible by $4$)
8. $a = 2$, $b = -34$
- $a + b = -32$ (even), $b - a = -36$ (divisible by $4$)
- $x^2 = -16$ (not a square), $y^2 = -9$ (not a square)
9. $a = 4$, $b = -17$
- $a + b = -13$ (odd), $b - a = -21$ (not divisible by $4$)
10. $a = 17$, $b = -4$
- $a + b = 13$ (odd), $b - a = -21$ (not divisible by $4$)
11. $a = 34$, $b = -2$
- $a + b = 32$ (even), $b - a = -36$ (divisible by $4$)
- $x^2 = 16$, $y^2 = -9$ (not a square)
12. $a = 68$, $b = -1$
- $a + b = 67$ (odd), $b - a = -69$ (not divisible by $4$)

So, the only valid case is $a = -2$, $b = 34$:
$$x^2 = 16 \implies x = \pm 4$$
$$y^2 = 9 \implies y = \pm 3$$

Check in the original equation:
$$x^4 + 68 = 4y^4$$
$$256 + 68 = 4 \times 81$$
$$324 = 324$$

Thus, all integer solutions are:
$$(x, y) = (4, 3), (4, -3), (-4, 3), (-4, -3)$$