Conjecture. For any real number $x \in (0,1]$ there exists a unique expansion in the form $x=-2+\sqrt{a_1+\sqrt{a_2+\sqrt{a_3+\cdots}}}$ with $a_k$ being natural numbers from the set $(2,3,4,5,6)$. This expansion always converges to $x$.
We can make a simple continued fraction expansion of any real number in a unified way, using a variation of Euclidean algorithm. For irrational numbers this expansion would be infinite. Sometimes it has a pattern (quadratic irrationals, $e$), most times it doesn't.
For quadratic irrationals, however, we can also make an infinite nested radical expansion, which is connected to its continued fraction expansion in a clear way, take Golden ratio for example.
$$\phi=\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cdots}}}=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}}$$
But in principle, we can make an infinite nested radical expansion for any real number. The 'best' rule for this is not clear, however, this seems doable:
$$\pi^2=9.8696044010893586188344909998761511353136994072408=7+p_1$$
$$p_1^2=8.234629418751416572757458690438995355335793971314=6+p_2$$
$$p_2^2=4.99356863914929388187773600296403392057375875115=2+p_3$$
$$\dots$$
$$\pi=\sqrt{7+\sqrt{6+\sqrt{2+\sqrt{6+\sqrt{6+\sqrt{5+\cdots}}}}}}$$
Or, using the notation I just invented (similar to continued fraction notation):
$$\pi=\sqrt{7+}\sqrt{6+}\sqrt{2+}\sqrt{6+}\sqrt{6+}\sqrt{5+}\sqrt{5+}\sqrt{2+}\sqrt{4+}\sqrt{6+}\sqrt{3+}\sqrt{4+}\sqrt{2+}\sqrt{4+}\sqrt{6+} \cdots$$
Now, why am I always taking $2$ as a whole part of the number I square? That's because if I take $1$ I quickly get a number in the form $1. \cdots$, so I can't go on further. And if I take $3$ I get bigger numbers under the root. So this seems optimal. In a similar fashion:
$$e=\sqrt{5+}\sqrt{3+}\sqrt{5+}\sqrt{3+}\sqrt{3+}\sqrt{6+}\sqrt{4+}\sqrt{4+}\sqrt{2+}\sqrt{4+}\sqrt{3+}\sqrt{2+}\sqrt{4+}\sqrt{4+}\sqrt{5+} \cdots$$
Here are plots for the convergence of above expansions for $\pi$ and $e$. You can see that the convergence is very good.
In fact, there are some interesting general properties of this expansion. If we only consider the numbers with the whole part $2$, then we get only numbers from $2$ to $6$ under the radical, with their mean value being around $4$.
How do I prove the observed properties of this expansion? Would this expansion always converge to the number? Is it possible that we get a pattern for some transcendental number in this expansion?
This is related to Bolyai Expansion, as was suggested by @Watson in a comment, however every $a_k$ here is not zero, which makes the expansion more 'orderly' in my opinion.
Also, I conjecture that taking $2$ as a whole part to square is the best way in the sense that if we take $1$ we may get $0$ as the next $a_k$, but if we take $3$ or more the set of $a_k$ becomes much larger, than just $(2,3,4,5,6)$.
Some more examples (I will use a more simple notations for the expansion as a set of numbers):
$$R(e)=(5,3,5,3,3,6,4,4,2,4,3,2,4,4,5,3,2,6,4,5,3,3,2,2,5,4,4,3,2,6,6,5,3,5,2,6,6,6,3,3,5,3,2,3,6,2,6,5,2,2,\dots)$$
$$R(\pi-1)=(2,4,5,2,6,6,4,6,4,4,5,4,6,6,5,2,2,3,5,5,2,3,3,2,5,2,6,2,5,3,2,2,5,4,3,2,2,4,6,3,4,3,3,6,3,3,5,2,2,6,\dots)$$
$$R(\gamma+2)=(4,4,6,6,3,3,4,2,2,2,4,2,5,2,4,5,3,2,2,4,4,5,5,6,2,4,6,3,3,5,3,3,5,3,2,2,5,2,6,4,2,3,4,3,4,6,6,4,6,2,\dots)$$
$$R(\phi+1)=(4,6,2,4,5,5,4,5,6,3,2,3,3,2,3,5,4,2,2,2,2,5,2,2,4,6,3,2,5,6,2,6,2,4,6,6,3,4,6,3,2,6,6,4,3,3,2,2,4,4, \dots)$$
We recover the well known identities and some other similar expansions:
$$R(2)=(2,2,2,2, \dots)$$
$$R(3)=(6,6,6,6,, \dots)$$
$$R(1+\sqrt{2})=(3,6,2,2,2,2, \dots)$$
$$R(\sqrt{5})=(3,2,2,2,2,2, \dots)$$
For some rational numbers (and quadratic irrationals) this expansion seems random:
$$R(2.5)=(4,3,2,3,2,3,3,5,6,6,5,4,2,6,6,2,6,4,2,2,2,4,2,4,5,2,5,5,2,2,6,2,6,3,4,2,3,6,2,4,4,3,2,3,2,2,5,4,3,2, \dots)$$
$$R(1+\sqrt{3})=(5,4,2,3,3,2,3,4,2,6,5,3,4,6,3,3,4,6,5,5,3,4,4,5,5,2,4,5,4,5,3,6,2,2,3,4,6,4,4,5,3,4,2,3,3,4,3,5,6,2, \dots)$$
By the way, this makes a beautiful (if inaccurate) approximations to some constants (see above):
$$\gamma \approx \sqrt{4+\sqrt{4+\sqrt{6+\sqrt{6+\sqrt{3+\sqrt{3}}}}}}-2=0.577$$
$$e \approx \sqrt{5+\sqrt{3+\sqrt{5+\sqrt{3+\sqrt{5}}}}}=2.718$$
And (cheating a little):
$$\pi \approx \sqrt{7+\sqrt{6+\sqrt{5}}}=3.1416$$