Infiniteness of irregular primes $\equiv3\pmod4$

Question

I want to prove there are infinitely many irregular primes of the form $4n+3$. I already know that

1. for prime $q$ such that $2q+1$ is composite there exists prime $p\equiv3\pmod4$ such that $p$ divides numerator of $B_{2q}$

2. for primes $p_s>\cdots>p_1>3$ and integer $q\equiv1\pmod{M}, M=(p_1-1)\cdots(p_s-1)$ none of those primes divides numerator of $B_{2q}/(2q)$ [see here]

3. if $p-1\nmid m$ then $B_m/m$ is $p$-integral (i do not know, if this is necessary here)

So let $p_1,\ldots,p_s$ be all irregular numbers of the form $4n+3$. Set $r=3kM+1$ (thus $2r+1$ is composite) and choose $k$ large enough to have $|B_{2r}/(2r)|>1$. Then there is a prime $p'$ with $v_{p'}(B_{2r}/(2r))>0$ and $p'-1\nmid 2r$ by Claussen–von Staudt. Also there is $p$ same as in (1).

I don't know how to carry on. I suppose I want to "identify" $p$ and $p'$. But what else?