I assume the "usual proof" goes by defining the obvious chain of elements $p_\eta$ by induction on $\eta$, and then arguing that that constitutes a chain with no upper bound.
I'm not sure if this counts, but: let $<$ be a well-ordering of the set of chains through $\mathbb{P}$. Now we can define a sequence of chains $C_\eta$ as: $C_\eta$ is the $<$-least chain such that $$\forall \beta<\eta, C_\beta\subsetneq C_\eta.$$ If we assume that $\mathbb{P}$ satisfies the hypothesis of Zorn's lemma but has no maximal elements - equivalently, that every chain $C$ through $\mathbb{P}$ has an upper bound not contained in $C$ - then $C_\eta$ exists for every ordinal $\eta$, which is a clear contradiction once we let $\eta$ exceed the cardinality of the set of chains.
Of course this still hinges on a transfinite induction/recursion argument - as Asaf comments, transfinite recursion is basically the only technique we have for utilizing well-orderings, so I don't think there's going to be a proof which avoids it. But notice that in this proof these arguments are pretty easy:
The recursion defining $C_\eta$ is very short.
To argue that $C_\eta$ always exists, only the limit case is at all interesting, and even then we only need the fact that $\bigcup_{\beta<\eta} C_\beta$ is a chain.
Additionally, to me at least this seems more direct than the usual one - in particular, we never need dirty our hands with individual elements of $\mathbb{P}$.
I'm not sure this satisfies the OP, though.