You don't need to know that $5 = (2-i)(2+i)$ is a prime factorization; all you need is that the two factors are not units.

One way to see that $2-i$ is not a unit is by computation:

$$ \mathbb{Z}[i] / (2-i) \xrightarrow{i \to x} \mathbb{Z}[x] / (x^2 + 1, 2-x) \xrightarrow{x \to 2} \mathbb{Z} / (2^2 + 1) \cong \mathbb{Z} / 5 $$

(all arrows are isomorphisms). The result isn't the zero ring, so $2-i$ is not a unit. The fact the result is a domain does additionally prove that $2-i$ is prime, though.

You don't need to know that $5 = (2-i)(2+i)$ is a prime factorization; all you need is that the two factors are not units.

One way to see that $2-i$ is not a unit is by computation:

$$ \mathbb{Z}[i] / (2-i) \xrightarrow{i \to x} \mathbb{Z}[x] / (x^2 + 1, 2-x) \xrightarrow{x \to 2} \mathbb{Z} / (2^2 + 1) \cong \mathbb{Z} / 5\Bbb Z $$

(all arrows are isomorphisms). The result isn't the zero ring, so $2-i$ is not a unit. The fact the result is a domain does additionally prove that $2-i$ is prime, though.