Why language is not regular
Yes, you are absolutely right.  GeeksforGeeks is just wrong in this case.

Your analysis is correct.  The language $L=\{a^n b^{2m} \mid n \ge 1, m \ge 1\}$ certainly is regular.  Your DFA seems fine.  I don't know why the GeeksforGeeks article claims it is not; but in any case they are wrong.  It's a strange example; even by their own notions of what counts as a "regular pattern", $b^{2m}$ is a "regular pattern", so I don't know why they are giving that example.

Even if we try to look at better examples, the claim is still wrong.  For instance, consider the language

$$L = \{a^n a^p \mid n \ge 0, p \text{ is prime}\}.$$

This language is described by the concatenation of a "regular pattern" and a "non-regular pattern", so that article claims it should be not regular -- but they're wrong.  In fact, $L$ is regular, as $L  = \{a^m \mid m \ge 2\}$.

This whole notion of a "regular pattern" and "non-regular pattern" in that article is pretty much baloney.  I realize what they're trying to do -- they're trying to look for a shortcut to quickly tell whether a language is regular based on the syntax of the description of the language -- but those proposed rules are not a reliable way to tell whether a language is regular or not, and the notion of a "regular pattern" is not well-defined.  That's not a standard concept -- it's just something someone on the Internet made up.  At best, it is a way to form a quick guess at whether a language is regular or non-regular -- but I would recommend always following that up by proving your guess is correct, as their proposed rules are a bit dodgy.

Perhaps a lesson is to rely on primary sources, like well-regarded textbooks, and don't trust too much what you read on the Internet.