This question is an offshoot of this earlier MSE question.
Let $\sigma(z)$ denote the sum of divisors of $z \in \mathbb{N}$, the set of positive integers. Denote the abundancy index of $z$ by $I(z) := \sigma(z)/z$.
If $N={p^k}{m^2}$ is an odd perfect number with Euler prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$, then it is somewhat trivial to prove that
$$3 - \frac{p - 2}{p(p-1)} = \frac{3p^2 - 4p + 2}{p(p-1)} < I(p^k) + I(m^2)$$
and
$$I(p^k) + I(m^2) \leq \frac{3p^2 + 2p + 1}{p(p+1)} = 3 - \frac{p - 1}{p(p+1)}.$$
Now, setting $x := 3 - \bigg(I(p^k) + I(m^2)\bigg)$, we have the simultaneous inequalities
$$\frac{p-1}{p(p+1)} \leq x < \frac{p - 2}{p(p - 1)}$$
resulting in the inequalities
$$\bigg((p - 2) > xp(p-1)\bigg) \land \bigg((p - 1) \leq xp(p+1)\bigg).$$
Notice that it is known that
$$\frac{57}{20} < I(p^k)+I(m^2) < 3$$
so that we know that
$$0 < x < \frac{3}{20}.$$
We now solve the inequalities one by one.
Solution is
$$\frac{(x+1) - \sqrt{x^2 - 6x + 1}}{2x} < p < \frac{(x+1) + \sqrt{x^2 - 6x + 1}}{2x}$$
Solution is
$$p \in \bigg(-\infty, \frac{1 - x - \sqrt{x^2 - 6x + 1}}{2x}\bigg] \bigcup \bigg[\frac{1 - x + \sqrt{x^2 - 6x + 1}}{2x},\infty\bigg)$$
Now, this is where the computations start to get messy. Can I ask for some help?
Basically, I would like to get numerical (lower [and upper?]) bounds for $p$.
It is easily seen that
$$\frac{(x+1) - \sqrt{x^2 - 6x + 1}}{2x} < 2, \frac{1 - x - \sqrt{x^2 - 6x + 1}}{2x} < 2$$
so your conditions reduces to
$$\frac{1 - x + \sqrt{x^2 - 6x + 1}}{2x} \leq p < \frac{1+x + \sqrt{x^2 - 6x + 1}}{2x}$$
then, one can see that the lower bound and upper bound differs by $1$ and both are unbounded in the given range of $x$. That is, there can be only one prime $p$ for any given value $x$.
Related Paper: The abundancy index of divisors of odd perfect numbers - Part III, http://nntdm.net/volume-23-2017/number-3/53-59/
