Wei's fourth-power q-supercongruence for n congruent to 3 modulo 4

Let $n$ be a positive integer with $n\equiv3\pmod4$, let $[n]=(1-q^n)/(1-q)$, let $(a;q)_k$ be the $q$-shifted factorial, and let $(a_1,\ldots,a_r;q)_k=\prod_{i=1}^r(a_i;q)_k$. Set $M=(n-1)/2$ or $M=n-1$. Wei's conjecture.

$$\sum_{k=0}^{M}(-1)^k[4k+1]\frac{(q;q^2)_k^4(q^2;q^4)_k}{(q^2;q^2)_k^4(q^4;q^4)_k}q^k\equiv[n]^2\frac{(q^3;q^4)_{(n-1)/2}}{(q^5;q^4)_{(n-1)/2}}q^{(1-n)/2}\pmod{[n]\Phi_n(q)^4}.$$

In particular, for $p\equiv3\pmod4$, the source proposes

$$\sum_{k=0}^{(p^r-1)/2}(-1)^k(4k+1)\frac{(\frac{1}{2})_k^5}{k!^5}\equiv p^{2r}\frac{(\frac{3}{4})_{(p^r-1)/2}}{(\frac{5}{4})_{(p^r-1)/2}}\pmod{p^{r+4}}.$$

This is motivated by numerical calculations and strengthens the earlier cubic-power $q$-congruence; the source gives no proof or resolution evidence.