Unique half rainbows
Charcoal, 31 bytes
Ｎθ≔ＥＸ⊕θ²Φ⊕ι∧λ¬∨›Ｘλ²ι﹪ιλη⁼¹№η§ηθ

Try it online! Link is to verbose version of code. Inefficiently outputs a Charcoal boolean, i.e. - for unique, nothing if not. Explanation:
Ｎθ

Input n.
≔ＥＸ⊕θ²Φ⊕ι∧λ¬∨›Ｘλ²ι﹪ιλη

For all integers up to (n+1)², list those factors that do not exceed its square root.
⁼¹№η§ηθ

See whether the nth list is unique.
Edit: Switching from (2n)² to (n+1)² made the code an order of magnitude faster without generating false positives. At a cost of 3 bytes, 2+⁴√n⁵ would be even faster; this works because the worst case scenario is of the form p₁²p₂ where p₁&lt;p₂&lt;p₁²; the next integer with the same half rainbow is p₁p₂², and since p₂&lt;p₁² we have p₂³&lt;p₁⁶ and therefore (p₁p₂²)⁴&lt;(p₁²p₂)⁵.