" the only non-field class of examples of UFDs having every nonzero element as a power of a fixed nonzero nonunit element is a (rank 1) discrete valuation ring."
Ok, but the UFD criterion is redundant here because there are no non-fields for which every nonzero element is a power of a fixed nonzero nonunit element. So that doesn't make UFDs any more relevant than any other extra criterion you could add.
Well, if you're willing to allow a bit of wiggle room in the form of "up to associates" (which is a qualifier that I forgot to add), there are integral domains in which every nonzero element equals the product of a unit and a power of a fixed nonzero nonunit element. This qualifier isn't too much to ask since "up to associates" is the generalized version of "up to a plus or minus" when factoring integers (which has no real bearing on factorizations).
In particular, let F be any field, and consider F[[x]], the ring of formal power series[1] over F. Given any nonzero f(x) in F[[x]], we can "factor out as many x's as we can" to write f(x) = x^n g(x) where n is a nonnegative integer, g(x) is in F[[x]], and the constant term of g(x) is nonzero (here we explicitly define x^0 to be 1). Thus, g(x) is a unit[2] (ie invertible), and every nonzero element of F[[x]] is a unit multiple of a power of x.
Ok, but rings for which every nonzero element is a power of a fixed nonzero nonunit element up to associates are exactly discrete valuation rings. So your statement boils down to "The only UFDs that are discrete valuation rings are (rank 1) discrete valuation rings". And I'm not sure what you mean by rank 1 here, but I can deduce that all discrete valuation rings are rank 1 from what you've said, since all discrete valuation rings are UFDs.
So what you've said boils down to:
"The only UFDs that are discrete valuation rings are discrete valuation rings", which again doesn't make UFDs more relevant than any other condition you might add.
Correction: the _integral domains_ for which every nonzero element is a power of a fixed nonzero nonunit element up to associates are exactly discrete valuation rings.
So my final statement should read:
"...which again doesn't make UFDs more relevant than any other condition you might add _that is stronger than the much more general condition of being an integral domain_"
"there are no non-fields for which every nonzero element is a power of a fixed nonzero nonunit element".
There are enough qualifiers here to go quite a bit further: Not only are there no non-fields of this sort, but, in fact, there are no rings of this sort whatsoever: if x is our fixed nonzero, nonunit element, we must have x + 1 as some power of x (x + 1 being nonzero, as x is not -1, as x is not a unit). Furthermore, this power cannot be zero (as x + 1 does not equal x^0 = 1, as x is nonzero). Thus, we would have x + 1 = x^(m + 1) for some m. Which is to say, 1 = x * (x^m - 1). But this makes x a unit, contra stipulation. Thus, there is no such ring.
[Of course, another way of looking at this is that, dropping the "nonzero nonunit" stipulation, we prove that the generating element x is either zero or a unit, and thus the rings with this property are, as noted, necessarily fields (or the trivial ring in which 0 = 1)]
It's seems a bit absurd to me to go further than non-fields, because fields have no non-zero non-units by definition.
That is a nicer argument than the one I had come up with though, which used the observation that all such rings must be quotients of the polynomial ring Z[x].
Ok, but the UFD criterion is redundant here because there are no non-fields for which every nonzero element is a power of a fixed nonzero nonunit element. So that doesn't make UFDs any more relevant than any other extra criterion you could add.