Prove a^4n b^m is irregular using puming lemma
Yes, you can use the pumping lemma. It looks like your line of thoughts just had a minor glitch.

Suppose your adversary claims the pumping lemma for $L$ works for some number $p\ge1$.
You can pick $z=a^{4(p+1)}b^p$, a word that "barely" satisfies the requirement.
Your adversary claims $z=uvw$ where $|v|\ge1$, $|uv|\le p$ and for all $n\geq 0$, $uv^{n}w\in L$.

Since $|uv|\le p&lt;4(p+1)$ and $z$ starts with $4(p+1)$ consecutive $a$'s, there must be no $b$ in $uv$. That is, $v=a^i$ for some $i=|v|\ge1$.

Consider $uv^0w=uw$. Since $z$ "barely" satisfies the requirement, removing some $a$'s without removing any $b$ will make $z$ not satisfying the requirement. You have beaten your adversary, again.

The above proof works, in fact, for any language $M$ such that $L\subseteq M\subseteq \{w\mid \#_a(w)\ge4(\#_b(w)+1)\}$.

I have been saying "$z$ 'barely' satisfies the requirement", since it is easier to understand that way although it is not very rigorous. It should be quite easy for readers to make it rigorous.