The formula for the lower bound of pi involving nested square roots looks like this: $p_{2^m} = 2^m\sqrt{2-\sqrt{2+\sqrt{2+ \sqrt{2+...}}}}$ where there are $m-1$ nested square roots.
For example, inscribing a square in a unit circle has $m = 2$, so the perimeter $p_4 = 2^2\sqrt2 = 4\sqrt2 < 2\pi$. Inscribing an octagon would have $m = 3$, so $p_8 = 2^3\sqrt{2-\sqrt2}$.
Deriving this formula, which seems to have significant relation to Viete's formula, involves inscribing a figure inside a circle and using various theorems related to circles and chords. I will upload the page to give some context.
I have been attempting to derive a formula for the upper bound, involving circumscribing figures about a unit circle. This seems to present more challenge to me, and I would greatly appreciate a step in the right direction. Please no trigonometry or angles, as the expression for pi must only involve these nested square roots.
I think most of my difficulty lies in sketching a decent figure, as it is not that geometrically obvious what a circumscribed $s_8$ would look like given a circumscribed $s_4$.
If $\theta$ is the half-angle subtened by each side of the polygon, you have
$\cos\theta=(1/2)\sqrt{2+\sqrt{2+\sqrt{2+...}}}.$
For comparison the factor in your lower-bound expression is
$\sin\theta=(1/2)\sqrt{2\color{blue}{-}\sqrt{2+\sqrt{2+...}}},$
which differs only by the sign indicated in blue. You should see that the squares of the sine and cosine properly add up to $1$.
The factor you want in the upper bound expression is $\tan\theta$, which is the second of these expressions divided by the first.
So, take your lower bound, multiply in one more power of $2$, and divide by the appropriate iteration of $\sqrt{2+\sqrt{2+\sqrt{2+...}}}$ (same length as the radical in the lower bound expression). Thus for the octagon you have the lower bound
$\pi_{-,8}=2^2\sqrt{2-\sqrt2}\approx3.061,$
and so the corresponding upper bound would be
$\pi_{+,8}=\dfrac{2^3\sqrt{2-\sqrt2}}{\sqrt{2+\sqrt2}}\approx3.314.$