Calculate 2nd order filter with gain
Here's your diagram with some annotations:

(I'm not promising anything about stability here. This is a high pass filter with gain significantly greater than 1. So this is only an idealized analysis, by the numbers. That's all. Not a practical recipe.)
Using sympy:
eq1 = Eq( va/(1/s/c) + va/r1 + va/(1/s/c), vo/r1 + vb/(1/s/c) + vi/(1/s/c) )
eq2 = Eq( vb/r2 + vb/(1/s/c), va/(1/s/c) )
eq3 = Eq( vc/r4 + vc/r3, vo/r3 )
eq4 = Eq( vo/r3 + vo/r1, va/r1 + vc/r3 + io )
ans = solve( [eq1, eq2, eq3, eq4, Eq( vb, vc )], [io, va, vb, vc, vo] )
simplify( ans[vo]/vi )

    c**2*r1*r2*s**2*(r3 + r4)/(c**2*r1*r2*r4*s**2 + 2*c*r1*r4*s - c*r2*r3*s + r4)

Moving that towards the standard high-pass form leads to the following:
$$\begin{align*}
\omega_{_0}&amp;=\frac1{C\sqrt{R_1\,R_2}}\\\\
\zeta&amp;=\sqrt{\frac{R_1}{R_2}}-\frac12\frac{R_3}{R_4}\sqrt{\frac{R_2}{R_1}}\\\\
K&amp;=1+\frac{R_3}{R_4}\\\\
\mathcal{H}\left(s\right)&amp;=K\frac{s^2}{s^2+2\zeta\,\omega_{_0}s+\omega_{_0}^{\:\!2}}
\end{align*}$$
I'll leave the algebraic details for you to find on your own.
Now, you want to make a Butterworth filter. The 2nd order normalized (also known as analytic) Butterworth constants are: \$s^2 + \sqrt{2} s + 1\$. (This is where \$\omega_{_0}=1\$.) So it follows that \$2\zeta=\sqrt{2}\$ or \$\zeta=\frac12\sqrt{2}\$. So you have this new requirement:
$$\sqrt{\frac{R_1}{R_2}}-\frac12\frac{R_3}{R_4}\sqrt{\frac{R_2}{R_1}}=\frac12\sqrt{2}$$
From here, it is advisable to set \$R_1=\eta\, R\$ and \$R_2=\frac{R}{\eta}\$ so that \$\omega_{_0}=\frac1{R\,C}\$. This gives you some freedom to adjust these two resistor values. Also, given your desired gain of \$A_v=K=2\$, it follows that \$\frac{R_3}{R_4}=1\$ and that:
$$\begin{align*}
\eta-\frac12\frac1{\eta}&amp;=\frac12\sqrt{2}\\\\
2\eta-\frac1{\eta}&amp;=\sqrt{2}\\\\
\therefore \eta&amp;=\frac14\left(\sqrt{2}+\sqrt{10}\right)\approx 1.14412281
\end{align*}$$
From here you can compute \$R_1\$ and \$R_2\$, given a desired \$R\$ used in setting \$\omega_{_0}\$.
Let's say \$\omega_{_0}=6283.18531\$ (\$f_{_0}=1\:\text{kHz}\$) and that \$C=100\:\text{nF}\$, per your starting condition. Then it follows that \$R\approx 1.592\:\text{k}\Omega\$ and therefore that \$R_1\approx 1.821\:\text{k}\Omega\$ and \$R_2\approx 1.391\:\text{k}\Omega\$.
My apologies for all that precision, here. But I'm retaining it so you can compare with the results from this recommended site to compute filter component values.
Using the E96 series of resistors, the above site says:
R1 = 1.82kΩ
R2 = 1.4kΩ
R3 = 10kΩ
R4 = 10kΩ
C1 = 0.1uF
C2 = 0.1uF

Which is obviously quite close to what I just computed.
Also, feel free to go there and look at the plot, as well.
Again. This is a high pass filter with a voltage gain you've specified as \$A_v=2\$! Isolated and by itself, I'm not at all suggesting that you try it with a high performance opamp. But computationally, at least, you can see where the numbers come from. (Or, at least, after you do some algebra.)
P.S. You can also choose to set \$R_1\$ and \$R_2\$ to the same value. But if so, and if keeping with the Butterworth response, then you will find that the gain is no longer 2. It will have to be less. Something closer to \$3-\sqrt{2}\approx 1.586\$.