The idea is to try to find a polynomial equality of the form $$X+\lambda P^2Q^2+\mu PR^2+\varepsilon Q^2R^2=\theta P^2QR \;\;\;(1)$$ with $\lambda$, $\mu$, $\varepsilon$, $\theta$ in $\mathbb{Q}$ and $P$, $Q$, $R$ in $\mathbb{Q}[X] $.
Let $A=X$, $B=\lambda P^2Q^2$, $C=\mu PR^2$ and $D=\varepsilon Q^2R^2$, then $$ABCD(A+B+C+D)=\lambda\mu\varepsilon\theta X(PQR)^5 \;.$$ Taking $X=kat^5$ with $k\in\mathbb{Q}$ such that $k\lambda\mu\varepsilon\theta=K^5$ with $K\in\mathbb{Q }$, we have $$wxyz(w+x+y+z)=a \;,$$ with $w=\dfrac{A}{KtPQR}$, $x=\dfrac{B}{KtPQR}$, $w=\dfrac{C}{KtPQR}$ and $w=\dfrac{D}{KtPQR }$.
In fact, we will see that we can find an equality of the form $(1)$ with $P$, $Q$ and $R$ of degree 1.
Take $P=X-\alpha$, $Q=X-\beta$ and $R=X-\gamma$ with $\alpha$, $\beta$ and $\gamma$ in $\mathbb{Q}$ (such that $\gamma\neq\beta$).
Suppose $(1)$ holds ; by replacing $X$ by $\alpha$, $\beta$ then $\gamma$, by derivating then by replacing $X$ by $\alpha$, and by considering the dominant terms, we obtain
$$\left\{\begin{array}{l}\alpha+\varepsilon(\alpha-\beta)^2(\alpha-\gamma)^2=0\\\beta+\mu(\beta-\alpha)(\beta-\gamma)^2=0\\\gamma+\lambda(\gamma-\alpha)^2(\gamma-\beta)^2=0\\1+\mu(\alpha-\gamma)^2+2\varepsilon(\alpha-\beta)(\alpha-\gamma)(2\alpha-\gamma-\beta)=0\\\theta=\lambda+\varepsilon\\\end{array}\right.$$ which is equivalent to $$\left\{\begin{array}{l} \varepsilon=-\dfrac{\alpha}{(\alpha-\beta)^2(\alpha-\gamma)^2}\\ \mu=-\dfrac{\beta}{(\beta-\alpha)(\beta-\gamma)^2}\\ \lambda=-\dfrac{\gamma}{(\gamma-\alpha)^2(\gamma-\beta)^2}\\ (\alpha-\beta)(\alpha-\gamma)(\beta-\gamma)^2+\beta(\alpha-\gamma)^3-2\alpha(\beta-\gamma)^2(2\alpha-\gamma-\beta) \;\;\;(2)\\ \theta=\lambda+\varepsilon\\ \end{array}\right.$$ Let $u=\dfrac{\alpha-\gamma}{\gamma-\beta}$; the identity $(2)$ gives $$(\alpha-\beta)u(\gamma-\beta)^3+\beta u^3(\gamma-\beta)^3-2\alpha(\gamma-\beta)^2(2( \alpha-\gamma)+\gamma-\beta) \;,$$ which is equivalent to $$(\alpha-\beta)u+\beta u^3-2\alpha(2u+1)=0 \;,$$ which leads to $$\left\{\begin{array}{l} \beta=\dfrac{3u+2}{u^3-u}\alpha\\ \gamma=\dfrac{u^2+3u+1}{(1+u)(u^2-1)}\alpha\\ \varepsilon=-\dfrac{u^2(u+1)^6(u-1)^4}{\alpha^3(u^3-4u-2)^4}\\ \mu=\dfrac{u^2(u+1)^4(u-1)^2(3u+2)}{\alpha^2(u^3-4u-2)^3}\\ \lambda=-\dfrac{u^2(u+1)^6(u-1)^3(u^2+3u+1)}{\alpha^3(u^3-4u-2)^4}\\ \theta=-\dfrac{u^3(u+1)^6(u-1)^3(u+4)}{\alpha^3(u^3-4u-2)^4}\\ \end{array}\right.$$ Conversely, one can verify that if $\beta$, $\gamma$, $\varepsilon$, $\mu$, $\lambda$ and $\theta$ are defined in terms of $u$ and $\alpha$ by the previous relations, then the polynomial relation $(1)$ is satisfied.
Taking $k=-\dfrac{\alpha u}{(u+1)^2(u-1)^2(u^2+3u+1)(u+4)(3u+2)}$, we have $k\lambda\mu\varepsilon\theta=K^5$ with $K=\dfrac{u^2(u+1)^4(u-1)^2}{\alpha^2(u^ 4-4u-2)^3}$; we then choose $X=kat^5$. Let's set $$\left\{\begin{array}{l} F(u)=(u+1)^2(u-1)^2(u^2+3u+1)(u+4)(3u+2)\\ G(u)=(u+1)(u-1)(u^2+3u+1)(u+4)(3u+2)^2\\ H(u)=(u-1)(u^2+3u+1)^2(u+4)(3u+2)\\ J(u)=(u+4)(3u+2)(u^3-4u-2)\\ \end{array}\right.$$ we obtain (after simplifications) the following $2-$parameterization of an infinite subset of solutions of $(E_a)$: $$\left\{\begin{array}{l} w=\dfrac{a(u-1)^2(u^2+3u+1)^2(u^3-4u-2)J(u)^2t^4}{(aut^5+F(u))(au^2t^5+G(u))(aut^5+H(u))}\\ x=\dfrac{(aut^5+F(u))(au^2t^5+G(u))}{u(u-1)J(u)t(aut^5+H(u))}\\ y=\dfrac{u(3u+2)(aut^5+H(u))}{t(au^2t^5+G(u))}\\ z=\dfrac{(au^2t^5+G(u))(aut^5+H(u))}{u(u^2+3u+1)J(u)t(aut^5+F(u))}\\ \end{array}\right.$$ By choosing for example $a=1$ and $u=3$, we obtain the following $1-$parametrization of an infinite subset of solutions of $(E_1)$ $$\left\{\begin{array}{l} w=\dfrac{18809562772t^4}{(3t^5+55594)(3t^5+93632)(9t^5+128744)}\\ x=\dfrac{(3t^5+93632)(9t^5+128744)}{6006t(3t^5+55594)}\\ y=\dfrac{33(3t^5+55594)}{t(9t^5+128744)}\\ z=\dfrac{(3t^5+55594)(9t^5+128744)}{57057t(3t^5+93632)}\\ \end{array}\right.$$