Proving $L = \{a^nb^m \mid n, m≥0, n \neq m\}$ is not regular by use of Pumping Lemma
This is a trick question. Your proof doesn't quite work, but can be turned into a proof, as follows.

Let $p$ be the constant in the pumping lemma, and consider the word $a^pb^{p+p!} \in L$. According to the pumping lemma, it can be decomposes as $xyz$, where $|xy| \leq p$, $y \neq \epsilon$, and $xy^iz \in L$ for all $i \geq 0$. Since $|xy| \leq p$, we must have $y = a^q$ for some $q \leq p$; since $y \neq \epsilon$, we must have $q \geq 1$. Let $i = p!/q + 1$. Then you can check that $xy^iz = a^{p+p!} b^{p+p!} \notin L$, contradicting the pumping lemma.

Here are two more proofs. The first uses closure properties. If $L$ were regular then so would the following language be: $a^*b^* \setminus L = \{ a^nb^m : n = m \}$. However, this language is known to be non-regular.

Another proof uses Myhill–Nerode theory. Let us say that two words $x,y$ are incomparable if there exists a word $z$ such that $xz \in L$ but $yz \notin L$, or vice versa. In any DFA for $L$, we must have $\delta(q_0,x) \neq \delta(q_0,y)$ (why?). Therefore, if we can find an infinite collection of pairwise incomparable words, then the language is not regular (why?). In the case of $L$, such a collection consists of the words $a^i$ for all $i \geq 0$. Indeed, if $i \neq j$ then $a^ib^j \in L$ whereas $a^jb^j \notin L$, showing that $a^i,a^j$ are incomparable.