Equation if we write in the General form:
$$aX^2+bXY+cY^2=eZ^2+jZW+tW^2$$
If in this equation there any equivalent to a quadratic form in which the root is an integer.
$$q=\sqrt{b^2+4a(e+j+t-c)}$$
Then there are solutions. They can be written by making the replacement.
$$x=(b(2(e+j+t)-b)+4ac)s-(b+2a)(j+2t)k$$
$$y=(b^2+4c(e+j+t-a))s^2-2(b+2c)(j+2t)sk+(j^2+4t(a+b+c-e))k^2$$
Then decisions can be recorded and they are as follows:
$$X=(b-2(e+j+t-c)\pm{q})p^2+2(q((j+2t)k-(b+2c)s)\pm{x})pn+$$
$$+(((2(e+j+t-c)-b)\pm{q})y+2((j+2t)k-(b+2c)s)x)n^2$$
$$***$$
$$Y=(\pm{q}-(b+2a))p^2+2(q((j+2t)k-(2(e+j+t-a)-b)s)\pm{x})pn+$$
$$+(((b+2a)\pm{q})y+2((j+2t)k-(2(e+j+t-a)-b)s)x)n^2$$
$$***$$
$$Z=(\pm{q}-(b+2a))p^2+2(q((j+2t)k-(b+2c)s)\pm{x})pn+$$
$$+(((b+2a)\pm{q})y+2((j+2t)k-(b+2c)s)x)n^2$$
$$***$$
$$W=(\pm{q}-(b+2a))p^2+2(q((2(a+b+c-e)-j)k-(b+2c)s)\pm{x})pn+$$
$$+(((b+2a)\pm{q})y+2((2(a+b+c-e)-j)k-(b+2c)s)x)n^2$$
$p,n,k,s $ - integers asked us.
Your equation, $a^2 + b^2 = c^4$, is the same as $a^2 + b^2 = c^2$, except that $c$ has been replaced with $c^2$. Thus you can find solutions by choosing $m$ and $n$ so that
$$
\begin{align}
a&=2mn \\ b&=m^2-n^2 \\ c^2&=m^2+n^2.
\end{align}
$$
But how can you make sure that $c$ is an integer? Easy: that last equation, $c^2 = m^2 + n^2$ says that $m,n$ and $c$ also form a Pythagorean triple. Thus you can find solutions with two new parameters, $r$ and $s$ such that
$$
\begin{align}
m&=2rs \\ n&=r^2-s^2 \\ c&=r^2+s^2.
\end{align}
$$
Substituting these in the first set of equations to find $a$ and $b$, you have
$$
\begin{align}
a&=4rs(r^2-s^2) \\ b&=6r^2 s^2-r^4-s^4 \\ c&=r^2 + s^2.
\end{align}
$$
For example, choosing $r=2$ and $s=1$ gives you $a = 24$, $b=7$, $c=5$ and, indeed, $24^2 + 7^2 = 5^4$.
Best Answer
To simplify matters, let $k = l$ as suggested in the comments. Then the pair of equations,
$$mn(m^2+n^2)=uv(v^2+u^2)\tag1$$ $$mn(m^2-n^2)=uv(v^2-u^2)\tag2$$
where the second one is yours can be translated, respectively, into familiar forms,
$$(m - n)^4 + (u + v )^4 = (m + n )^4 + (u - v )^4\tag3$$ $$(m - n z)^4 + (u z + v )^4 = (m + n z)^4 + (u z - v)^4\tag4$$
with the imaginary unit $z=\sqrt{-1}$.
It is known that $(3)$ does not have a complete parameterization, so presumably likewise for $(4)$. However, starting with the solutions in the comments, then using elliptic curves, one can generate as many polynomial solutions to $(1)$ or $(2)$ as one wishes.