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Let $D$ be a square free integer. Let $R_D$ be the integral closure of $mathbb{Z}$ in the field $mathbb{Q}(sqrt{D})$.

For some values $D$, the ring $R_D$ is a $UFD$, but not for all. For example, the Gaussian integers $R_{-1}$ are a $UFD$ whereas the ring $R_{-5}$ is not. There are several ways to show this, including computing the class number of $R_D$. However, all the proofs I've seen feel ad hoc and unintuitive.

According to the Stark-Heegner theorem, for $D<0$, the ring $R_D$ is a $UFD$ if and only if
$$D in {-1,-2,-3,-7,-11,-19,-43,-67,-163}.$$
Is there any intuitive reason why this should be a complete list? Ideally there would be a structural reason - coming up with a separate proof for each $D$ in the list is deeply unsatisfying to me.

I wouldn't be so quick to discount the value of going through the list one by one. For this post, let's stipulate $D < 0$ throughout.

You've noticed $-2$ is the only even value on there, right? If $D$ is even, then $D = (sqrt D)^2$, which is obvious enough. But if $D$ is even and composite, it means that $N(z) = 2$ for $z in mathcal{O}_{mathbb{Q}(sqrt D)}$ is impossible. So then $2$ is irreducible, yet $D = 2 times x = (sqrt D)^2$, where $x in mathcal{O}_{mathbb{Q}(sqrt D)}$ also.

You've also noticed that $-5$ is not on the list either. Neither is $-13$, $-17$, $-29$, etc. What these numbers have in common, besides being odd, is that they are congruent to $3 bmod 4$ (remember that congruence gets "flipped" for negative numbers, so $-3 equiv 1 bmod 4$, not $3 bmod 4$).

So, if $D equiv 3 bmod 4$, then $N(1 + sqrt D) = -D + 1$, which is even. But in this domain, it turns out that $N(z) = 2$ is also impossible. Which means that $-D + 1$ has at least two distinct factorizations. Thus $D = -5$ gives us the classic example $6 = 2 times 3 = (1 - sqrt{-5})(1 + sqrt{-5})$.

Now let's say $D equiv 1 bmod 4$ instead. Then it's still the case that $-D + 1 = (1 - sqrt{D})(1 + sqrt{D})$, but... $$frac{1 - sqrt{D}}{2}, frac{1 + sqrt{D}}{2}$$ are also algebraic integers, both with minimal polynomial $$x^2 - x + frac{-D + 1}{4},$$ e.g., $$frac{1 - sqrt{-43}}{2}, frac{1 + sqrt{-43}}{2}$$ both have the polynomial $x^2 - x + 11$.

Therefore, the full factorization of $44$ in this domain is not $(1 - sqrt{-43})(1 + sqrt{-43})$ but $$2^2 left(frac{1 - sqrt{-43}}{2}right) left(frac{1 + sqrt{-43}}{2}right).$$

Why this doesn't work out for $D leq -167$ is quite a bit more involved, maybe someone else will address that.

I do think there is an intuitive reason, and it has to do with the number 41. You know that $n^2 + n + 41$ is prime for all $0 leq n < 41$? Of course it's composite for $n = 41$.

Notice that $$left(frac{1}{2} - frac{sqrt{-163}}{2}right)left(frac{1}{2} - frac{sqrt{-163}}{2}right) = 41,$$ $$left(frac{3}{2} - frac{sqrt{-163}}{2}right)left(frac{3}{2} - frac{sqrt{-163}}{2}right) = 43,$$ $$left(frac{5}{2} - frac{sqrt{-163}}{2}right)left(frac{5}{2} - frac{sqrt{-163}}{2}right) = 47,$$ and so on and so forth to $$left(frac{79}{2} - frac{sqrt{-163}}{2}right)left(frac{79}{2} - frac{sqrt{-163}}{2}right) = 1601.$$

What are the other numbers $d$ such that $n^2 + n + d$ gives primes for $0 leq n < d$? 2, 3, 5, 11, 17, see https://oeis.org/A014556 And then $-4d + 1$ gives $-7, -11, -19, -43, -67$, and $-163$ corresponds to 41.

No higher number $d$ meets this requirement, and thus does not correspond to a Heegner number.