Wolstenholme's theorem is stated as follows: if $p>3$ is a prime, then \begin{align*} \sum_{k=1}^{p-1}\frac{1}{k}\equiv 0 \pmod{p^2},\\ \sum_{k=1}^{p-1}\frac{1}{k^2} \equiv 0 \pmod{p}. \end{align*} It is also not hard to prove that $$ \sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 0 \pmod{p}. $$ However, there are some relationships between $\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}$ and $\sum_{k=1}^{p-1}\frac{1}{k^2}$ mod $p^2$, which I can not prove.

Question: If $p$ is an odd prime, then $$ 4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 3\sum_{k=1}^{p-1}\frac{1}{k^2}\pmod{p^2}. $$ I have verified this congruence for $p$ upto $7919$.

Comments:

(1) Since $$4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}=2\sum_{k=1}^{\frac{p-1}{2}}\frac{1}{k^2}-4\sum_{k=1}^{p-1}\frac{1}{k^2},$$ we need to prove $$2\sum_{k=1}^{\frac{p-1}{2}}\frac{1}{k^2}\equiv 7\sum_{k=1}^{p-1}\frac{1}{k^2} \pmod{p^2}.$$ This idea was given by Fedor Petrov.

(2) It is interesting that the congruence in the question is ture mod $p^3$ for $p\ge 7$, i.e., $$ 4\sum_{k=1}^{p-1}\frac{(-1)^k}{k^2}\equiv 3\sum_{k=1}^{p-1}\frac{1}{k^2}\pmod{p^3}, \quad \text{for $p\ge 7$}. $$ This was conjectured by tkr.

I appreciate any proofs, hints, or references!