For example, in $\mathbb{Z}[\sqrt{34}]$, it is clear that $\langle 2 \rangle = \langle 2, \sqrt{34} \rangle^2$ and $\langle 17 \rangle = \langle 17, \sqrt{34} \rangle^2$.
However, looking at the numbers rather than the ideals generated by them, I see that $2 = (6 - \sqrt{34})(6 + \sqrt{34})$ and $17 = (-1)(17 - 3 \sqrt{34})(17 + 3 \sqrt{34})$.
It seems clear to me from this that $\langle 2 \rangle = \langle 6 + \sqrt{34} \rangle^2$, since $6 - \sqrt{34} \in \langle 6 + \sqrt{34} \rangle$ by way of the unit $35 + 6 \sqrt{34}$ and thus $\langle 6 - \sqrt{34} \rangle = \langle 6 + \sqrt{34} \rangle$.
However, I'm not seeing how to prove that $17 - 3 \sqrt{34} \in \langle 17 + 3 \sqrt{34} \rangle$, and I'm also beginning to wonder if I'm even on the right track with this line of thought.
Yes, it's still a ramified ideal. But first, let's look at a prime number that really does split in this domain: $$(519 - 89 \sqrt{34})(519 + 89 \sqrt{34}) = 47.$$ Now try $$\frac{519 - 89 \sqrt{34}}{519 + 89 \sqrt{34}} = \frac{538675 - 92382 \sqrt{34}}{47}$$ and $$\frac{519 + 89 \sqrt{34}}{519 - 89 \sqrt{34}} = \frac{538675 + 92382 \sqrt{34}}{47}.$$ This tells us that neither factor of 47 is a divisor of the other.
By contrast, you have already observed that $$\frac{6 - \sqrt{34}}{6 + \sqrt{34}} = 35 - 6 \sqrt{34}$$ (which is a unit) and $$\frac{6 + \sqrt{34}}{6 - \sqrt{34}} = 35 + 6 \sqrt{34}$$ (which is also a unit). Daniel Fisher has done similar calculations for 17 in the comments.
Therefore $\langle 2 \rangle = \langle 6 + \sqrt{34} \rangle^2$ (as you've already noted) and $\langle 17 \rangle = \langle 17 + 3 \sqrt{34} \rangle^2$. The ideals ramify even though the generating numbers look like they split.
It's a similar situation with 5 and 11 in $\mathbb Z[\sqrt{55}]$. We can generalize: if $p \mid d$ and $p = (a - b \sqrt d)(a + b \sqrt d)$, then there is a unit $u$ such that $u(a - b \sqrt d) = a + b \sqrt d$ and vice-versa, and therefore $\langle p \rangle = \langle a + b \sqrt d \rangle^2$.