Infinitely many $n$ with $\tau(n+k)\ll k$ for all $k\geq 1$? (Erdős #826)

For a positive integer $m$ let $\tau(m)$ denote the number of divisors of $m$. Is there an absolute constant $C$ such that, for infinitely many $n$, one has $$\tau(n+k)\leq Ck\quad\text{for every }k\geq 1?$$ Equivalently: are there infinitely many $n$ for which $\tau(n+k)\ll k$ holds uniformly over all $k\geq 1$, with an implied constant independent of both $n$ and $k$?

Acceptance. FULLY RESOLVES: a complete proof (Lean/Coq preferred, otherwise a full written proof) that there is an absolute constant $C$ with $\tau(n+k)\leq Ck$ for all $k\geq 1$ for infinitely many $n$; OR a proof that no such $n$ exist (for every constant $C$, only finitely many $n$ satisfy $\tau(n+k)\leq Ck$ for all $k$). ADVANCES: reduce Lau's exponent — a proof that for infinitely many $n$, $\tau(n+k)\ll k^{C'}$ for all $k\geq 1$, with an explicit $C'$ strictly smaller than the exponent $C$ established in Lau's [La26] result (state Lau's exponent, whatever value it is proved at, as the bar to beat); or an unconditional counting/density statement on the set of admissible $n$. Deliver the proof (formal or written) with the explicit improved exponent.