Because 5.10 regular expressions have named captures you can use as rules, languages not being regular doesn't mean they cannot be matched by a Perl regexp. With named captures and rules, Perl regexes can match any context-free grammar. With backreferences, even more.
Now, I don't think the language `{1`^{n} | n = b^{c}, c > 1} is context-free, but that's harder to prove than it being non-regular. And it's still not sufficient to prove it cannot be matched by a Perl regexp without the use of `(?{ })` or `(??{ })`. | [reply] [d/l] [select] |

Any context-free language over a single-character alphabet is also regular (which can be shown using, for example, Parikh's theorem). So since this "language of exponentials" is non-regular, it is also non-context-free. Another way to look at it is that the pumping lemma for CFLs essentially collapses into the pumping lemma for regular languages when applied to single-character alphabets, because the concatenation operator is commutative on strings over a single-character alphabet.
I do stand by my intuition that backrefs alone (i.e., classical regexes + backrefs) won't help. But you are right, I am not able to prove it -- there is just no formal model that I'm aware of that exactly captures the expressivity of those operations, that would be amenable to impossibility proofs. I admit I hadn't thought of the new named captures & rules from 5.10 (I'm a bit behind the times). Clearly the named captures alone won't get you the regex desired in this case, but maybe some clever combination of both named rules & backrefs? I remain slightly skeptical but open-minded ;)
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