Let us first define the Euler polynomials to be the polynomials $P_n(q)$ that satisfy $$ \frac{qP_n(q)}{(1 - q)^{n+1}} = \Big(q\frac{d}{dq}\Big)^n\frac{q}{1 - q}. $$
For example, $P_0(q) = P_1(q) = 1$, and $P_2(q) = 1 + q$. One should note that these polynomials are always palindromic, and of degree $n - 1$, and satisfy a pretty easy-to-determine recursion from the above definition.
Next, let us consider the following expression. $$ T_n(q, x) = (1 - q)^{2n + 2} - x^2 q P_{2n+1}(q) $$ For $n = 0$, this is simply $$ (1 - q)^2 - x^2q = 1 - (x^2 + 2)q + q^2. $$ We note that this polynomial is also palindromic (as $(1 - q)^{2n + 2}$ is), and so we can factor it as $$ T_n(q, x) = \prod_{j=0}^n(1 - \zeta_j q)(1 - \zeta_j^{-1} q) $$ for some expressions $\zeta_j$.
The goal is now to write $x$ as a function of these $\zeta_j$. For example, if $n = 0$, then we have $$ 1 - (x^2 + 2)q + q^2 = (1 - \zeta q)(1 - \zeta^{-1} q) $$ or $$ x^2 = \zeta + \zeta^{-1} - 2 \iff x = \zeta^{1/2} - \zeta^{-1/2} $$ which is a nice and simple expression.
For higher $n$, let us define $\mu_j = \zeta_j^{1/2} - \zeta_j^{-1/2}$, and so $\mu_j^2 = \zeta_j + \zeta_j^{-1} - 2$. Then, for example, for $n = 1$ we have that $$ \begin{align} x^2 &= \mu_1^2 + \mu_2^2 \\ -6x^2 &= \mu_1^2\mu_2^2 \end{align} $$ and in general what we find is that for each $n$, that we have a series of expressions $$ c_{n, k} x^2 = \sigma_k(\mu_1^2, \ldots, \mu_n^2) $$ where $c_{n, k}$ are some integral coefficients (related to A036969) and $\sigma_k$ is the $k$-the elementary symmetric polynomial. These expressions should allow us to derive relations among the $\zeta_j$s, but it doesn't seem clear to me how these should look; in particular, for $n = 1$ the best that I can find is that $$ x = \frac{\mu_1^2}{\sqrt{6 + \mu_1^2}} = \frac{\zeta_1 + \zeta_1^{-1} - 2}{\sqrt{4 + \zeta_1 + \zeta_1^{-1}}} $$ which isn't really as nice as the expression $x = \zeta^{1/2} - \zeta^{-1/2}$.
As for motivation (e.g. why $x^2$?), what I'm working with is an expression of the form $$ \sum_n A_n(q)x^{2n} = \prod_m T_k(q^m,x) $$ and the right-side of this expression is manifestly (quasi-)modular; what I'd like to understand is what that means for the left-hand side.
