Find all triplets $(x, y, \ell) \in \mathbb{N}^{3}$ such that

$$
x^{3}+y^{3}-53=7^{\ell}
$$
If $l \geqslant 1$, we can look modulo 7. The equation rewrites as $x^{3}+y^{3}-4 \equiv 0$, which is $x^{3}+y^{3} \equiv 4$. However, the cubes modulo 7 are $-1,0$ and 1, so the sum of two cubes can never equal 4. Therefore, there is no solution for $l \geqslant 1$.

If now $l=0$, the equation rewrites as $x^{3}+y^{3}=54$. We can test the small cases $x=0,1,2$ and verify that there is no solution since $54-x^{3}\left(=y^{3}\right)$ is not the cube of an integer. For $x=3$, we have $y=3$ which is a solution. If $x \geqslant 4$, there is no solution since $x^{3}>54$ and $y^{3} \geqslant 0$.

Therefore, we have a unique solution: $(3,3,0)$.