Bound the length of a path along which an entire function outgrows every power $z^n$ (Erdős #514)

Let $f$ be a transcendental entire function and let $M(r)=\max_{\lvert z\rvert=r}\lvert f(z)\rvert$. Boas proved (unpublished) that there is a path $L$ to infinity such that, for every fixed $n$, $$\left\lvert\frac{f(z)}{z^{n}}\right\rvert\to\infty\qquad\text{as } z\to\infty \text{ along } L.$$ Can the length of such a path $L$ (say, up to modulus $r$) be estimated in terms of $M(r)$? More strongly, does there exist a path along which $\lvert f(z)\rvert$ tends to $\infty$ faster than a fixed function of $M(r)$, for instance faster than $M(r)^{\epsilon}$ for some fixed $\epsilon>0$?

Acceptance. FULLY RESOLVES: a complete proof (machine-checkable in Lean/Coq preferred, otherwise a full written proof) establishing (i) an explicit estimate for the minimal length of a path $L$ with $\lvert f(z)/z^n\rvert\to\infty$ (all $n$) in terms of $M(r)$, and (ii) a decisive answer to whether, for every transcendental entire $f$, some path carries $\lvert f(z)\rvert\ge M(r)^{\epsilon}$ for a fixed $\epsilon>0$ and all large $r$ — either by constructing such paths or by proving the $M(r)^{\epsilon}$ growth is in general unattainable. ADVANCES: prove a length bound for a natural subclass (e.g. entire functions of finite order, or of regular/completely regular growth), with proof; or exhibit a specific transcendental entire function for which every such path has length at least an explicit increasing function of $r$ (a lower bound), with proof; or establish the $M(r)^{\epsilon}$ growth along a path for a restricted class. Deliver the theorem and its proof, or the explicit counterexample function together with a verified path-length lower bound.