We define $v_{3}(n)$ to be the non-negative integer $k$ such that $3^{k} \mid n$ but $3^{k+1} \nmid n$. The equation is equivalent to

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

We will consider the cases $y \neq 2 x$ and $y=2 x$ separately.

Case 1. Suppose $y \neq 2 x$. Since $45^{x}>45^{x}-6^{y}=2019^{z}>45^{z}$, then $x>z$ and so $2 x>z$. We have

$$z=v_{3}\left(3^{z} \cdot 673^{z}\right)=v_{3}\left(3^{2 x} \cdot 5^{x}-3^{y} \cdot 2^{y}\right)=\min \{2 x, y\}$$

as $y \neq 2 x$. Since $2 x>z$, we get $z=y$. Hence the equation becomes $3^{2 x} \cdot 5^{x}-3^{y} \cdot 2^{y}=$ $3^{y} \cdot 673^{y}$, or equivalently,

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

Case 1.1. Suppose $y=1$. Doing easy manipulations we have

$$3^{2 x-1} \cdot 5^{x}=2+673=675=3^{3} \cdot 5^{2} \Longrightarrow 45^{x-2}=1 \Longrightarrow x=2$$

Hence one solution which satisfies the condition is $(x, y, z)=(2,1,1)$.

Case 1.2. Suppose $y \geqslant 2$. Using properties of congruences we have

$$1 \equiv 2^{y}+673^{y} \equiv 3^{2 x-y} \cdot 5^{y} \equiv(-1)^{2 x-y} \bmod 4$$

Hence $2 x-y$ is even, which implies that $y$ is even. Using this fact we have

$$0 \equiv 3^{2 x-y} \cdot 5^{y} \equiv 2^{y}+673^{y} \equiv 1+1 \equiv 2 \bmod 3$$

which is a contradiction.

Case 2. Suppose $y=2 x$. The equation becomes $3^{2 x} \cdot 5^{x}-3^{2 x} \cdot 2^{2 x}=3^{z} \cdot 673^{z}$, or equivalently,

$$5^{x}-4^{x}=3^{z-2 x} \cdot 673^{z}$$

Working modulo 3 we have

$$(-1)^{x}-1 \equiv 5^{x}-4^{x} \equiv 3^{z-2 x} \cdot 673^{z} \equiv 0 \bmod 3$$

hence $x$ is even, say $x=2 t$ for some positive integer $t$. The equation is now equivalent to

$$\left(5^{t}-4^{t}\right)\left(5^{t}+4^{t}\right)=3^{z-4 t} \cdot 673^{z}$$

It can be checked by hand that $t=1$ is not possible. For $t \geqslant 2$, since 3 and 673 are the only prime factors of the right hand side, and since, as it is easily checked $\operatorname{gcd}\left(5^{t}-4^{t}, 5^{t}+4^{t}\right)=1$ and $5^{t}-4^{t}>1$, the only way for this to happen is when $5^{t}-4^{t}=3^{z-4 t}$ and $5^{t}+4^{t}=673^{z}$ or $5^{t}-4^{t}=673^{z}$ and $5^{t}+4^{t}=3^{z-4 t}$. Adding together we have

$$2 \cdot 5^{t}=3^{z-4 t}+673^{z}$$

Working modulo 5 we have

$$0 \equiv 2 \cdot 5^{t} \equiv 3^{z-4 t}+673^{z} \equiv 3^{4 t} \cdot 3^{z-4 t}+3^{z} \equiv 2 \cdot 3^{z} \bmod 5$$

which is a contradiction. Hence the only solution which satisfies the equation is $(x, y, z)=$ $(2,1,1)$.

Alternative Solution by PSC. Working modulo 5 we see that $-1 \equiv 4^{z} \bmod 5$ and therefore $z$ is odd. Now working modulo 4 and using the fact that $z$ is odd we get that $1-2^{y} \equiv 3^{z} \equiv 3 \bmod 4$. This gives $y=1$. Now working modulo 9 we have $-6 \equiv 3^{z} \bmod 9$ which gives $z=1$. Now since $y=z=1$ we get $x=2$ and so $(2,1,1)$ is the unique solution.