Determine all natural numbers $x, y$ and $z$ such that

$$
45^{x}-6^{y}=2019^{z}
$$
We start by decomposing each term into a product of prime factors. The equation then becomes

$$
3^{2 x} \cdot 5^{x}-3^{y} \cdot 2^{y}=3^{z} \cdot 673^{z}
$$

We will now handle different cases:
$\triangleright$ If $y>2 x$, then $3^{z} \cdot 673^{z}=3^{2 x}\left(5^{x}-3^{y-2 x} \times 2^{y}\right)$. Gauss's theorem then indicates that $3^{z}$ divides $3^{2 x}$ (since $5^{x}-3^{y-2 x} \times 2^{y}$ is not divisible by 3, and is thus coprime with $3^{z}$) and that $3^{2 x}$ divides $3^{z}$ (since $673^{z}$ is not divisible by 3, and is thus coprime with $3^{2 x}$). We deduce that $z=2 x$, and the equation becomes

$$
673^{2 x}=5^{x}-3^{y-2 x} \cdot 2^{y}
$$

However, since $673^{2 x} \geqslant 5^{x}$, this equation has no solution, and this case is therefore impossible.
$\triangleright$ If $y=2 x$, then $3^{z} \cdot 673^{z}=3^{2 x}\left(5^{x}-2^{y}\right)$. As before, Gauss's theorem indicates that $3^{2 x}$ divides $3^{z}$. We deduce that $2 x \leqslant z$, and the equation becomes

$$
3^{z-2 x} \cdot 673^{z}=5^{x}-2^{y}
$$

However, since $673^{z} \geqslant 5^{z} \geqslant 5^{x}$, this equation has no solution either, and this case is therefore impossible.
$\triangleright$ If $y<2 x$, then $3^{z} \cdot 673^{z}=3^{y}\left(3^{2 x-y} \cdot 5^{x}-2^{y}\right)$. Once again, since neither $673^{z}$ nor $3^{2 x-y} \cdot 5^{x}-2^{y}$ are divisible by 3, Gauss's theorem indicates that $3^{z}$ and $3^{y}$ divide each other. We deduce that $z=y$, and the equation becomes

$$
673^{y}=3^{2 x-y} \cdot 5^{x}-2^{y}
$$

We then distinguish several sub-cases:
$\triangleright$ If $y=0$, the equation becomes $2=3^{2 x} \cdot 5^{x}$, and of course has no solution.
$\triangleright$ If $y=1$, the equation becomes $675=3^{2 x-1} \cdot 5^{x}$. Since $675=3^{3} \times 5^{2}$, we deduce the existence of the solution $(x, y, z)=(2,1,1)$.
$\triangleright$ If $y \geqslant 2$, consider the equation modulo 4: it becomes $1 \equiv 3^{y}(\bmod 4)$, which means that $y$ is even.
Then, since we necessarily have $x \geqslant 1$, and if we consider the equation modulo 5, it becomes $(-2)^{y} \equiv-2^{y}(\bmod 5)$, which means that $y$ is odd. The equation therefore has no solution in this sub-case.
The unique solution to the equation is thus the triplet $(2,1,1)$.