First Solution: The key to this solution is the identity
$$a^3 + b^3 - 1 + 3ab = (a+b-1)(a^2 + b^2 - ab + a + b + 1).$$
Even though this can be verified readily, we explain how to find it. We start with a polynomial of degree 3 which has $a$, $b$ and an auxiliary number $c$ as its roots $f(x) = (x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab+ac+bc)x - abc$. We obtain the identity
$$0 = f(a) + f(b) + f(c) = a^3 + b^3 + c^3 - 3abc - (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc).$$
Substituting $c = -1$ gives the identity mentioned above.

The original equation $a^3 + b^3 + 3ab = 53$ is equivalent to the equation
$$(a+b-1)(a^2+b^2-ab+a+b+1) = 52.$$
Because we are looking for integer solutions, the two numbers $u = a+b-1$ and $v = a^2+b^2-ab+a+b+1 = (a+b)^2+(a+b-1)-3ab+2 = (u+1)^2+u+2-3ab = (u+1)(u+2)+1-3ab$ have to be divisors of $52 = 2 \cdot 2 \cdot 13$. The integer divisors of $52$ are $\pm 1, \pm 2, \pm 4, \pm 13, \pm 26, \pm 52$. In particular, $u \equiv \pm 1 \pmod 3$. We have $a+b = u+1$ and $3ab = (u+1)(u+2)+1-v$. The last identity and $u \equiv \pm 1 \pmod 3$ imply $v \equiv 1 \pmod 3$. Hence, the following table covers all possibilities for $u$ and $v$.

| u | v | a+b | 3ab | ab |
|----|----|-----|-----|-----|
| 1 | 52 | 2 | -45 | -15 |
| -2 | -26| -1 | 27 | 9 |
| 4 | 13 | 5 | 18 | 6 |
| 13 | 4 | 14 | 207 | 69 |
| -26| -2 | -25 | 603 | 201 |
| 52 | 1 | 53 | 2862| 954 |

From $x = a+b$ and $y = ab$ we obtain $(a, b)$ by solving the quadratic equation $a^2 - ax + y = 0$ and using $b = x - a$. For $a$ to be an integer, we need therefore that the discriminant of the quadratic, $x^2 - 4y$, is a perfect square. Only the first and third row of the table give $x^2 - 4y \ge 0$. Therefore, the only possible solutions are $(5, -3), (-3, 5), (3, 2), (2, 3)$. Substituting them in the original equation shows that they are indeed solutions.

Second Solution: Because $53$ is not a perfect cube, neither $a$ nor $b$ could be zero. Without loss of generality, we may assume $a \le b$. We distinguish the following three cases.

**$0 < a \le b$:** In this case, $a^3 + b^3 + 3ab \ge 2a^3 + 3a^2 = a^2(2a + 3)$, hence $a^2(2a+3) \le 53$. This implies $a \le 2$. If $a = 1$, we get $b^3+3b = b(b^2+3) = 52 = 2 \cdot 2 \cdot 13$. As $b < b^2+3$, the only possibilities for $b$ are $1, 2, 4$ but none of them leads to a solution. If $a = 2$, we have to solve $b^3+6b = b(b^2+6) = 45 = 5 \cdot 3 \cdot 3$. As $b < b^2+6$, $b$ can only be $1, 3$ or $5$. Only $b = 3$ is a solution.

**$a \le b < 0$:** In this case, $a^3 + b^3 + 3ab = a(a^2 + 3b) + b^3 = 53 > 0$. Because $b^3 < 0$ and $a < 0$ this implies $a^2 + 3b < 0$. Hence, $a^2 < 3|b|$. On the other hand, $|a| \ge |b|$ and so $b^2 \le a^2 < 3|b|$ which implies $|b| < 3$. If $b = -1$, we get $a^3 - 3a = a(a^2 - 3) = 54 > 0$. As $a < 0$ this implies $a^2 < 3$, i.e. $a = -1$ which does not give a solution. If $b = -2$, we get $a^3 - 6a = a(a^2 - 6) = 61$. As before, this implies $a^2 < 6$, i.e. $a = -1$ or $a = -2$. But both values do not solve the equation.

a < 0 < b: In this case, $3ab < 0$ and so $a^3 + b^3 > 0$ which implies $b > |a|$. Let $c = -a > 0$ and $b = c + k$ with $k \ge 1$. The given equation becomes
$$\begin{aligned} b^3 - c^3 - 3bc &= 53 \\ (c + k)^3 - c^3 - 3c(c + k) &= 53 \\ 3c^2(k - 1) + 3ck(k - 1) + k^3 &= 53. \end{aligned}$$
The left hand side is at least $3(k-1)+3k(k-1)+k^3 = k^2(k+3)-3$, hence we need to have $k^2(k+3) \le 56$ which implies $k \le 3$. We cannot have $k=1$, because $53$ is not a perfect cube. If $k=2$ we need to solve $3c^2+6c+8=53$, or equivalently, $c(c+2) = 15$ with $c=3$ as its only positive solution. This leads to $(a,b) = (-3,5)$. Finally, if $k=3$ we need to have $6c^2+18c+27=53$ which has no solution as $53$ is not divisible by $3$.

Therefore, the only solutions to the original equation are $(5,-3), (-3,5), (3,2)$ and $(2,3)$.