We know that the Diophantine equation $x^2+y^4=2z^4$ has infinitely many solutions .
Some of them are shown below
$$(y,z)=(1,1),(1,13),(1343,1525),(2372159,2165017).$$
I investigated the ratio of $\dfrac{z}{y}$ from 64 solutions (are the smallest)
and found that $84.076$ is maximum.
Do we know that the ratio $\dfrac{z}{y}$ is bounded or not.
Claim: $\displaystyle\;\frac{z}{y}$ is unbounded.
Let $(x,y,z)$ stands for any integer solution for the equation
$$x^2 + y^4 = 2z^4,\;\; z > 0\tag{*1}$$ Factorize above equation over Gaussian integers which is an UFD, we can conclude there are integers $u,v$ and $0 \le \epsilon < 4$ such that
$$ x + iy^2 = i^\epsilon (1+i) (u+iv)^4\quad\text{ and }\quad z = u^2 + v^2 $$
For the purpose of proving $\displaystyle\;\frac{z}{y}\;$ is unbounded, we only need to consider the case $\epsilon = 0$.
From now on, we will assume $\epsilon = 0$. Let $$\begin{cases} \phi_x(u,v) &\stackrel{def}{=} u^4-4u^3v-6u^2v^2+4uv^3+v^4\\ \phi_{y^2}(u,v) &\stackrel{def}{=} u^4+4u^3v-6u^2v^2-4uv^3+v^4\\ \phi_z(u,v) &\stackrel{def}{=} u^2 + v^2 \end{cases} $$ we have $$x + iy^2 = \phi_x(u,v) + i \phi_{y^2}(u,v)\quad\text{ and }\quad z = \phi_z(u,v)\tag{*2}$$ Define another bunch of functions $$ \begin{cases} e(X,Y) &= Y^2 - (X^3 - 2X),\\ \psi_y(X,Y) &= X( 2XY - (X^3 - X^2 - 6X - 2))\\ \psi_x(X,Y) &= \phi_x(Y-X,X(X+1))\\ \psi_{y^2}(X,Y) &= \phi_{y^2}(Y-X,X(X+1))\\ \psi_z(X,Y) &= \phi_z(Y-X,X(X+1)) \end{cases} $$ By brute force, one can verify
$$\psi_{y^2}(X,Y) - \psi_y(X,Y)^2 = e(X,Y)(Y^2+4X^2Y-10x^4-23x^3-12x^2-2x)$$
For any $(X,Y) \in \mathbb{Q}$, we can associate with it a pair of integers $(u,v) \in \mathbb{Z}^2$ by the rule
$$\frac{u}{v} = \frac{Y-X}{X(X+1)}\quad\text{ and }\quad \gcd(u,v) = 1$$
When $e(X,Y) = 0$, we have $\psi_{y^2}(X,Y) = \psi_y(X,Y)^2$. This in turn implies
$$\phi_{y^2}(u,v) = \left(\frac{v}{X(1+X)}\right)^4 \psi_y^2(X,Y) = \left[\left(\frac{v}{X(1+X)}\right)^2 \psi_y(X,Y)\right]^2 $$ is the square of a rational number. Since $\phi_{y^2}(u,v)$ is an integer, it is the square of an integer!
As a result, any rational point $(X,Y)$ on the elliptic curve $$\mathcal{E} : \{\; (X,Y) \in \mathbb{R}^2 : e(X,Y) = 0\; \}$$ leads to an integer solution to the Diophantine equation $(*1)$ through the parametrion $(*2)$.
For such a solution, we have
$$\frac{z}{y} = \frac{\psi_z(X,Y)}{\sqrt{\psi_{y^2}(X,Y)}} = \left|\frac{\psi_z(X,Y)}{\psi_y(X,Y)}\right| = \left|\frac{2Y-(X^3+3X^2+2X-2)}{2XY-(X^3-X^2-6X-2)}\right| $$
As a consequence, if the rational points on $\mathcal{E}$ can get arbitrary close to the real solutions for following set of equations:
$$\begin{cases} Y^2 &= X^3 - 2X\\ 2XY &= X^3-X^2-6X-2 \end{cases} \tag{*3} $$
then $\displaystyle\;\frac{z}{y}\;$ is unbounded. This set of equations has five solutions, they lie approximately at $$ \begin{align} r_1 &= (-1, 1 ),\\ r_2 &= ( -0.9449472918334532, -1.022803578367951),\\ r_3 &= (-0.2813045676720525, 0.7350842599417211),\\ r_4 &= ( 2.116520167087263, -2.290904920616812),\\ r_5 &= ( 7.109731692418242, 18.57862423904303) \end{align} $$ It turns out the first three solutions are useless to us. We only need $r_4$ and $r_5$.
For the group of rational points on $\mathcal{E}$, it has the structure
$$(\mathbb{Z}/2\mathbb{Z})(0,0) \oplus \mathbb{Z}(2,2)$$
The Torsion subgroup $(\mathbb{Z}/2\mathbb{Z})(0,0)$ is useless to us. The remaining free part has rank $1$ and is generated by the generator $(2,2)$. Let $(X,Y)_k$ be the point on $\mathcal{E}$ corresponds to summing $(2,2)$ $k$ times using the addition on $\mathcal{E}$.
Consider the Weierstrass elliptic function $\wp(z)$ satisfying the ODE
$$\wp'(t)^2 = 4\wp(t)^3 - 8\wp(t)$$
One can parametrize the complex version of our elliptic curve $\mathcal{E}$ by
$$\mathbb{C} \in t \quad\mapsto\quad (X,Y) = \left(\wp(t),\frac12\wp'(t)\right) \in \mathcal{E}_{\mathbb{C}} = \big\{\; (X,Y) \in \mathbb{C}^2 : e(X,Y) = 0 \;\big\}$$
It is known that $\wp(t)$ has fundamental half periods $\omega_1, \omega_2 = i\omega_1$ where$\color{blue}{^{[1]}}$ $$2\omega_1 = \frac{\Gamma\left(\frac14\right)^2}{2^{7/4}\sqrt{\pi}} \approx 2.204878797993166$$
When we move $t$ from $-\omega_1$ to $\omega_1$ along the real axis, we know
for $t \in [-\omega_1,0)$, $\wp(t)$ increases from $\sqrt{2}$ to $+\infty$ while $\wp'(t)$ increases from $0$ to $+\infty$.
for $t \in (0,\omega_1]$, $\wp(t)$ decreases from $+\infty$ to $\sqrt{2}$ while $\wp'(t)$ increases from $-\infty$ to $0$.
This means for any $(X,Y) \in \mathcal{E}$ with $X > \sqrt{2}$, the corresponding $t$ falls on the real axis. In particular, this happens to the generator $(2,2)$ and the real solutions $r_4, r_5$ for equation $(*3)$.
Let $T_g$, $T_4$ and $T_5$ be the parameters corresponding to $(2,2)$, $r_4$ and $r_5$ respectively. Above discussion allow us to pick them such that$\color{blue}{^{[2]}}$
$$0 < T_4 < \omega_1 < T_g < T_5 < 2\omega_1 \quad\implies\quad \begin{cases} T_4 \approx 0.7258941119069,\\ T_g \approx 1.451788223813796,\\ T_5 \approx 1.828333510903481 \end{cases} $$
Define the sequence $\displaystyle\;t_k = 2\omega_1 \left\lfloor\frac{ k T_g }{2 \omega_1}\right\rfloor$.
Since $(2,2)$ is a generator for the free part of the group, $\frac{T_g}{2\omega_1}$ is irrational, the set $\left\{ t_k : k \in \mathbb{Z}_{+}\right\}$ will be dense in $[0,2\omega_1]$. This implies we can find two subsequences $\alpha_k$, $\beta_k$ such that
$$\lim_{k\to\infty} t_{\alpha_k} = T_4 \quad\text{ and }\quad \lim_{k\to\infty} t_{\beta_k} = T_5 $$ In terms of corresponding rational points on $\mathcal{E}$, we have $$ \lim_{k\to\infty} (X,Y)_{\alpha_k} = r_4 \quad\text{ and }\quad \lim_{k\to\infty} (X,Y)_{\beta_k} = r_5 $$ This justifies our claim that $\displaystyle\;\frac{z}{y}$ is unbounded.
As an illustration how this works, consider following values of $t_k$.
$$\begin{cases} t_{21} &\approx 1.824128326178556\\ t_{122} &\approx 0.7278594658298088\\ t_{264} &\approx 1.828059034024375 \end{cases} \quad\text{ versus }\quad \frac{z}{y} = \begin{cases} 84.07559131784495, & k = 21,\\ 179.9151342283949, & k = 122\\ 1282.614986769345, & k = 264 \end{cases} $$
As one can see, $t_{21}$ and $t_{264}$ is close to $T_4$ (especially $t_{264}$) while $t_{122}$ is close to $T_5$, this explains why the corresponding $\frac{z}{y}$ is high and the observed behavior that as a function of $k$, the size of $\frac{z}{y}$ is sort of unpredictable.
Notes
$\color{blue}{[1]}$ The elliptic function we used here is essentially a scaled version of Lemniscatic elliptic function, a lot of properties like the half-periods can be derived from it.
$\color{blue}{[2]}$ Given the numerical values of $\;(\wp(t),\wp'(t)) = (X,2Y)$, we can compute the corresponding parameter $t$ using the function
InverserWeierstrassP[{ X, 2*Y }, {8,0}]on wolfram alpha.