A cubic q-supercongruence with $(q;q^2)_k$ factors

Let $m$ and $n$ be positive integers with $n$ odd, and let $(a;q)_k$ denote the q-shifted factorial. The cubic q-supercongruence conjecture.

$$\sum_{k=0}^{mn-1}(-1)^k[3k+1]\frac{(q;q^2)_k^3}{(q;q)_k^3} \equiv (-1)^{(n-1)/2}q^{(n-1)^2/4}[n]\sum_{k=0}^{m-1}(-1)^k(3k+1)\frac{8^k(\frac12)_k^3}{k!^3}\;\allowbreak(\operatorname{mod}\Phi_n(q)^2).$$

The source records this as another simplified version of an earlier conjecture; no resolution is supplied.