Proving $a^{2+n}a^{n}$ is regular using the Pumping Lemma
Noooooo!

Usage of the pumping lemma to prove regularity is remarkably rare, and certainly won't happen with a direct application.

Let's review the formal statement, with particular emphasis on the quantifiers:

$$\forall \ L \subseteq \Sigma^* ( \ L \ \text{is regular} \Rightarrow  \exists \ p \ge 1 \ \ \forall \ w \dots \ \ \exists \ x, y, z \dots \ \ \forall \ i \dots \ )$$

Two notable observations are that being regular is part of the hypotheses and that nothing is said about the converse, which is in fact false (a number of classical counterexamples can be found in literature).

In other words, even if you managed to somehow prove the $\exists \forall \exists \forall$ block, it still wouldn't say anything about the regularity of $L$. The way the pumping lemma is normally used is in its contrapositive form. You may recall from basic logic that $(p \rightarrow q) \rightarrow (\neg q \rightarrow \neg p)$. Using that principle, we can derive the following logical consequence:

$$\forall \ L \subseteq \Sigma^* ( \ \forall \ p \ge 1 \ \ \exists \ w \dots \ \ \forall \ x, y, z \dots \ \ \exists \ i \dots \Rightarrow L \ \text{is }\textbf{not }\text{regular} \ )$$

Notice how this is a tool to prove non-regularity, and that, even there, nothing can be said about the converse.

I'm afraid your attempt of proof is just entirely misguided.