Why, after squaring, is the smallest root of the quartic the solution of the real problem? Looking at this post, I considered the more general problem of the solution of equation $$x+\sqrt{x+\sqrt{a+x}}=a\tag 1$$ where $a> 0$ .
Using successive squaring processes, this leads to the quartic 
$$x^4-2 (2
   a+1) x^3+\left(6 a^2+4 a+1\right) x^2-\left(4 a^3+2 a^2+1\right) x+a \left(a^3-1\right)=0\tag2$$ the discriminant of which being
$$\Delta=1024 a^4-1280 a^3-768 a^2-80 a-23$$ which is positive for any $a>\sqrt 3$; then, using the notations used here, since $P=-4(4a+1)$ and $D=-32(4a+1)$, all four roots are real and distinct. Extensive numerical tests showed that all roots are positive.

How could be explained the fact that the only solution of $(1)$ systematically corresponds to the smallest root of $(2)$ ? 
How could be explained the fact that, for a few integer values of $a$, one of the solution of $(2)$ is also an integer ? Testing for the range $2 \leq a \leq 1000$, this is the case for the values $a=2,3,5,11,46,68,250,296,566,636$.


If the real and positive solutions from the 4th-order equations are $x_1 < x_2 < x_3 < x_4$ and we introduce
$$
y_i^{(\alpha,\beta)} = x_i + \alpha \sqrt{x_i + \beta \sqrt{x_i + a}}
$$
where $\alpha,\beta = \pm$ and only take the positive roots. For each $(\alpha,\beta)$ there is a unique $x_i$ that will result in $y_i^{(\alpha,\beta)}=a$. 
We can now make two observations:


*

*From the ordered $x_i$ it automatically follows that:
$$
y_1^{(++)} < y_2^{(++)} <y_3^{(++)} <y_4^{(++)}
$$ 

*For the different possibilities of $x_1$ we have
$$
\max \left( y_1^{(--)}, y_1^{(-+)}, y_1^{(+-)}, y_1^{(++)} \right) = y_1^{(++)} 
$$
Hence if the solution of $x + \sqrt{x + \sqrt{x+ a}} =a$ would not correspond to $x_1$ (the smallest) but to $x_j$ one would obtain that $y_1^{(\alpha,\beta)} \leq y_1^{(++)} < a=y_j$, which contradicts the fact that there should be such a combination $(\alpha,\beta)$.
With respect to the solutions you missed the case $a=250$ and $a=296$ for values $a \leq 500$.
We have the following set of equations in integers at the different levels of square roots:
$$
x + a = n^2
$$
$$
x + n = m^2
$$
$$
x + m = a
$$
for some integers $n,m \in \mathbb{Z}$. From the second we obtain $n=m^2-x$ and substitution of this and the third equation in the first gives:
$$
x + (x+m) = (m^2 - x)^2
$$
and hence
$$
x^2 - 2(m^2+1)x + m(m^3 -1) = 0
$$
With two solutions:
$$
x_\pm = (m^2+1) \pm \sqrt{2 m^2 + m + 1}
$$
and the corresponding values for $a$ are:
$$
a_\pm = (m^2+m+1) \pm \sqrt{2 m^2 + m + 1}
$$
In order for this to be integers we only need to find value of $m$ (positive and negative) such that $2 m^2 + m + 1$ is a perfect square.
So the problem simplifies to finding integer solutions of the equation:
$$
2 m^2 + m + 1 = p^2
$$
This can be rewritten as
$$
8 p^2 - (4 m + 1)^2 = 7
$$
which is a type of Pell equation. The succesive solutions of $m_i$ that results in such a perfect square are given by $m_0=0$, $m_1=1$ and the recurrence relation $m_{i+1}=-6 m_{i} - m_{i-1} - 2$. Note that in order to obtain all solutions $(x,a)$ of the problem above, both positive and negative values of $m$ are allowed and that the recurrence relation needs to be followed in both directions. 
With the substitution $m_i \rightarrow \frac{1}{4} (u_i - 1)$, a normal recurrence relation is obtained with $u_0=0$, $u_1=5$, and $u_{i+1}=-6 u_{i} - u_{i-1}$. Working this out properly gives:
$$
m_i = \frac{1 - 2 \sqrt{2}}{8} \left( -3 - 2 \sqrt{2} \right)^i + \frac{1 + 2 \sqrt{2}}{8} \left( -3 + 2 \sqrt{2} \right)^i - \frac{1}{4}
$$
and results in 
$$
\begin{array}{llllll}
m_0=0 & m_1=1     & m_2=-8    & m_3=45     & m_4=-264   & \dots \\
      & m_{-1}=-3 & m_{-2}=16 & m_{-3}=-95 & m_{-4}=552 & \dots
\end{array}
$$
Inserting this sequence in the expressions for the solutions $(x,a)$ gives:
$$
(x_+,a_+)=\dots,(280,296),(14,11),(2,2),(4,5),(76,68),\dots
$$
$$
(x_-,a_-)=\dots,(234,250),(6,3),(0,0),(0,1),(54,46),\dots
$$
which are all integer combinations for which $x + a  - ((a-x)^2 - x)^2 == 0$. Only a subset of them will also give $x + \sqrt{x + \sqrt{x+ a}} =a$. 
Note that also the $(x_i,a_i)$ could be written in a similar form, but more terms, as done for $m_i$.
