The general criterion is that no number can be found with more than one valid, distinct factorization. This might sound like I'm merely rephrasing the question, but it's actually a reframing of the question.
Plenty of numbers (infinitely many, to be precise) in $\mathbb{Z}[\sqrt{-5}]$ have more than one factorization. $6$ is just the easiest to find. To oversimplify matters, your main concern is with the "natural" primes from 2 to $p < 4|d|$ or $p \leq d$ as needed.
Now, the case of $\mathbb{Z}[\sqrt{2}]$ is actually more complicated than you might realize. Part of the complication is that $\sqrt{2}$ is a real number and so $\mathbb{Z}[\sqrt{2}]$ has infinitely many units. This sets up the trap of infinitely many factorizations that are not distinct because they involve multiplication by units, e.g., $$7 = (3 - \sqrt{2})(3 + \sqrt{2}) = (-1)(1 - 2\sqrt{2})(1 + 2\sqrt{2}) = (5 - 3\sqrt{2})(5 + 3\sqrt{2}) = \ldots$$
But $7$ really does have only one distinct factorization in $\mathbb{Z}[\sqrt{2}]$, as you can see by dividing these numbers by $1 + \sqrt{2}$, and $\mathbb{Z}[\sqrt{2}]$ really is a UFD. But the full explanation may require me to make several assumptions about what you know.
Let's look at a "simpler" domain, $\mathbb{Z}[\sqrt{10}]$, though it certainly has some of the same traps as $\mathbb{Z}[\sqrt{2}]$: $$31 = (-1)(3 - 2\sqrt{10})(3 + 2\sqrt{10}) = (11 - 3\sqrt{10})(11 + 3\sqrt{10}) = (-1)(63 - 20\sqrt{10})(63 \ldots$$
You have to look at numbers that are already composite in $\mathbb{Z}$ to begin with. And if $d = pq$, where $p$ and $q$ are distinct primes, the choice of where to look first is obvious: $$10 = 2 \times 5 = (\sqrt{10})^2.$$
Verify that $$\frac{\sqrt{10}}{2} \not\in \mathbb{Z}[\sqrt{10}], \frac{\sqrt{10}}{5} \not\in \mathbb{Z}[\sqrt{10}], \frac{2}{\sqrt{10}} \not\in \mathbb{Z}[\sqrt{10}], \frac{5}{\sqrt{10}} \not\in \mathbb{Z}[\sqrt{10}].$$ This means that $\mathbb{Z}[\sqrt{10}]$ is not UFD and we didn't need to compute any logarithms or sines to come to this conclusion. (You're starting to see why integral closure matters in making these determinations, right?)
Contrast $\mathbb{Z}[\sqrt{6}]$: $$6 = (2 - \sqrt{6})(2 + \sqrt{6})(3 - \sqrt{6})(3 + \sqrt{6}) = (\sqrt{6})^2$$ but $$\frac{\sqrt{6}}{2 + \sqrt{6}} = 3 - \sqrt{6}$$ and so on and so forth. This means that $6 = (\sqrt{6})^2$ is an incomplete factorization, just as, say, $81 = 9^2$ is in $\mathbb{Z}$. But this is not enough to prove that $\mathbb{Z}[\sqrt{6}]$ is or is not UFD.
As it turns out, both $\mathbb{Z}[\sqrt{2}]$ and $\mathbb{Z}[\sqrt{6}]$ are UFDs, and what is probably the simplest, most common way of proving this requires a full understanding of ideals. Adapting the proof that $\mathbb{Z}$ is a UFD to these domains can be done, but that has its own pitfalls.
"$\Rightarrow$" If $\mathfrak p$ is minimal over $(a)$, then $\mathfrak p$ contains a prime element $p$ which appears in the decomposition of $a$ into primes. It follows that $\mathfrak p=(p)$.
"$\Leftarrow$" (I suppose that $R$ is an integral domain.) Use the Kaplansky's Theorem for UFDs: a domain is a UFD iff every nonzero prime ideal contains a nonzero prime. Let $\mathfrak q$ be a non-zero prime ideal, and $a\in\mathfrak q$, $a\ne 0$. Then there is $\mathfrak p$ a prime ideal contained in $\mathfrak q$ which is minimal over $(a)$. Since $\mathfrak p$ is principal, then it is generated by a prime element, so $\mathfrak q$ contains a nonzero prime.
Best Answer
It is enough to show that $\mathbb{Z}[2\sqrt{2}]$ is not a unique factorisation domain (why?).
The elements $2$ and $2\sqrt{2}$ are irreducible and $$ 8 = (2\sqrt{2})^2 = 2^3, $$ so the factorisation is not unique.