In a quadratic ring $\mathbb{Z}[\sqrt{d}]$ that is not a UFD, is there a simple proof that if an element has a norm that is a prime integer, then that element of the ring is prime, not merely irreducible? [It would be sufficient to show that it is impossible for two elements which are neither conjugates nor associates to have the same prime norm, but I can't think of a simple proof of that, either.]
For example, in $\mathbb{Z}[\sqrt{-5}]$, a well-known non-UFD, where norms $a^2+5b^2$ can only be $\equiv1$ or $\equiv9$ mod $20$ to be prime in $\mathbb{Z}$, prime elements are $3+2\sqrt{-5}$, $6+1\sqrt{-5}$, $4+3\sqrt{-5}$, $\ldots$, with respective norms $29,41,61,\ldots$; and certainly the numbers $29,41,61,\ldots$ can only be expressed uniquely in the form $a^2+5b^2$. But is there a simple proof that these primes have a unique expression in that form, in a way that could easily be generalised to other quadratic rings?
A proof that does not require understanding of ideals in rings would be very much appreciated, but maybe that is too much to ask!
