Yep, this looks correct. Nice job!

You've shown that $m^2\neq p(pq^2+2pq+p)$, but it could still be the case that $m^2=pk$ for some other integer $k$.

@AlannRosas So, how to handle it, Sir?

@SharkyKesa Really? What about the implication, Sir?

Why don't you want to use Euclid's lemma? It's a pretty elementary lemma with an elementary proof.

@morrowmh Because it doesn't studied yet in the class (although I already study it myself).

Okay, then maybe you can use the fundamental theorem of arithmetic instead, see math.stackexchange.com/questions/3061922/…

The obvious problem with your proof of the lemma is that you don't use the fact that $p$ is prime. This means that if your proof was valid it would work for non-prime $p$ which is clearly false.

@AlannRosas I think, it is very clear. Since I have shown that $m^2<p$, it follows that $p \nmid m^2$, for example, if $36<p$, pick $p=37$ and then $37 \nmid 6^2$(?).

@Fishbane So, how to handle it, Sir?

Showing $m^2<p(pq2+2pq+p)$ is not the same as showing $m^2<p$. Easier is to use the fundamental theorem of arithmetic to show that if $p$ divides $m^2$ then $p$ divides $m$.

@PitikKepit $m^2<p(pq^2+2pq+p)$ does not imply $m^2<p$.

@PitikKepit Oh sorry, I misread one of your lines. You're basically done once you assume $m = pq + r$. What you have is $p \mid p^2 q^2 + 2pqr + r^2$, so $p \mid r^2$. Since $0 < r < p$, $p \nmid r^2$ so we have a contradiction.

@PitikKepit There are a lot of ways you could prove it. Without knowing why you are excluding certain proofs I can't give any solid recommendations. As others have said FTA is a pretty good idea. It is also going to depend on how you have defined a prime, in some contexts the fact that $p|ab\implies p|a \text{ or } p|b$ is the definition of a prime so the result is trivial.

Many thanks in advanced to you all, Sir.

@Sharky How do you know that $0<r<p$ implies $p\nmid r^2$ without using Euclid's lemma (or something equivalent to it)?

@Sharky Well sure, it's easy if we can assume FTA. Euclid's lemma is easier than FTA

@PM2Ring We are trying to avoid Euclid's Lemma, so I think this is fine? I don't know how else we can avoid Euclid's Lemma.

Sorry to ask, Sir. How to prove it without using FTA too? Is it so hard? I mean, how to prove the lemma just using the inequality $0<r<p$?

As @Fishbane already commented, you're not using the fact that $p$ is prime, so something must be wrong. The problem is here: "But, since $0 < r < p$, we have $p \not\mid r^2$, a contradiction." This statement needs explanation; it uses the assumption that $p$ is prime.

@Magdiragdag Yes, Sir. How to conclude that if $0<r<p$ and $p$ prime, then $p \nmid r^2$?

Couple of errors I came across: 1. $p| r^2$ does not follow from anything you have said prior to that, Notice $5 | 15$ but $5 \not | 12+3$. Then you said $0<r<p$ so $p \not | r^2$, again this is wrong. Notice, $0<3<9$ and $9\not | 3$ but $9|3^2$. Everything you said after that is clearly wrong too.

Well, it's your proof. Note that your using the implication $p \mid x^2 \implies p \mid x$ at other places in your proof as well. It follows, of course, from the fact that $p$ is prime. Using that here would work, but then you might as well jump from $p \mid m^2$ to $p \mid m$ right away without the detour through $r$.