Acknowledging that $\mathbb{Z[\mathcal{i}]}$ is a UFD, show that $11 + 14i, 23 − 12i, −91 + 230i, 305 + 192i$ cannot all be prime.
By direct calculation, I checked that $$(11+14i)(305+192i)=(23-12i)(-91+230i)$$ now if all four numbers were prime, then they'd be irreducible, so as $\mathbb{Z[\mathcal{i}]}$ is a UFD, there'd be a permutation $\sigma \in S_2$ such that $a_{i}$ and $b_{\sigma(i)}$ would be associates, but as the only units in $\mathbb{Z[\mathcal{i}]}$ are $\pm 1, \pm \mathcal{i}$, we can check that the elements in the left hand are not associates of those in the right hand, so they cannot be all prime. Is this a valid idea?
Yes, your idea is good and works.
Another common approach to check if $a+bi$ is prime is to compute is norm $$N(a+bi)=a^2+b^2$$
The norm is multiplicative meaning that $N(\alpha\beta)=N(\alpha)N(\beta)$. The nice thing is that this implies that $a+bi$ is prime if $N(a+bi)=a^2+b^2=p$ for some prime $p$. We can then check:
Proof: ($\Rightarrow$) Since $\alpha$ is a unit there exists $\beta\in\Bbb C$ such that $\alpha\beta=1$. Taking norms we get $$N(\alpha\beta)=N(\alpha)N(\beta)=1$$ This forces $N(\alpha)=\pm 1$, but $N(\alpha)\geq 0$ so $N(\alpha)=1$.
($\Leftarrow$) We may write $$N(\alpha)=\alpha\bar\alpha=1$$ where $\bar\alpha$ is the complex conjugate of $\alpha$. $\overline{a+bi}=a-bi$. Thus $\alpha$ is a unit.
Proof: Suppose $\alpha=\beta\gamma$, taking norms we have $$N(\alpha)=N(\beta)N(\gamma)=p$$ Thus either $N(\beta)=1$ or $N(\gamma)=1$, so either $\beta$ or $\gamma$ is a unit. Hence $\alpha$ is a prime.