I come across this question very long ago. I just got one solution by my computer search. If any one know the other solutions and resolvability, please let me know.
$$x^4 + 16x^2y^2 + y^4 = z^2$$ has a solution $(1,2,\color{brown}9)$. (Typo corrected.)
Can we find others?
I seem to remember proving that for there to be a solution there needed to be a smaller one. Using the given one as that gave new solutions. I can't find my work on it though, if any member can...Please discuss
The trouble with writing this answer is that the "give a man a fish" part is so much more elementary than the "teach him to fish part". But I'll try to do both.
First, the easy remark. Set $X=x/y$ and $Z=z/y^2$. If $x^4+16x^2y^2+y^4=z^2$, then $X^4+16X^2+1 = Z^2$ with $X$ and $Z$ rational. Conversely, if we have a solution to $X^4+16X^2+1 = Z^2$ with $X$ and $Z$ rational then, for $d$ large enough, $(Xd, d, Zd^2)$ is a solution to $x^4+16x^2y^2+y^4=z^2$ in integers. So we can search for rational solutions to $X^4+16 X^2+1 =Z^2$. Rationals are easier to deal with than integers, and we have one fewer variable to keep track of.
The Fish:
Set $$U = Z - X^2 - 8 \quad V=X(Z-X^2-8).$$ Notice that $$X = V/U \quad Z = U + (V/U)^2 + 8.$$ So $(U,V)$ are rational if and only if $(X,Z)$ are.
Rewrite our given equation $$Z^2 - (X^2+8)^2 = -63$$ $$(Z-X^2-8) (Z+X^2+8) = -63.$$ $$U (U+2(V/U)^2+16) = -63.$$ $$U^3 + 2 V^2 + 16 U^2 = - 63U.$$ $$2 V^2 = - U^3 - 16U^2 -63 U.$$
Every step is reversible, except that the solution $(U,V) = (0,0)$ does not correspond to a choice of $(X,Z)$. So we come down to finding rational points on $2 V^2 = - U^3 - 16 U^2 - 63 U$.
Now, there is software for finding rational points on $V^2 = \mathrm{cubic}$. I'm not very skilled at this sort of thing, so I'm hoping some other reader will do that part. This answer is CW to make it easy to edit in the result.
How to fish:
This part of the answer assumes some familiarity with working with algebraic curves. I've tried to write this in an elementary way, but, if you haven't see any of this, you might like to work through a book like Fulton's Algebraic Curves first.
So, where did the substitution $(U,V) = (Z-X^2-8, X(Z-X^2-8))$ come from? What I knew was that the curve $Z^2 = X^4+16X^2+1$ is a genus one curve with two punctures. That means that the complex solutions to this equation look like an inner tube with two holes punched in it. Near one puncture, $X$ and $Z$ are both large and $Z \approx X^2$, near the other puncture $X$ and $Z$ are both large and $Z \approx -X^2$. Software to work with elliptic curves wants the curve in the form $V^2 = \mathrm{cubic}$, which is a genus one curve with one puncture.
Algebraically, I need to compute the ring of polynomials in $X$ and $Z$ which don't blow up at one of the two punctures. I chose to work near the puncture where $Z \approx X^2$.
Being more precise, $$Z = \sqrt{X^4+16x^2+1} = X^2 \sqrt{1+16/X^2+1/X^4} = X^2 (1+8 X^{-2} - (63/2) X^{-4} + 256 X^{-6} + \cdots )$$ $$=X^2+8 - (63/2) X^{-2} + 256 X^{-4} + \cdots.$$ (Series expansion found by the same method my calculus students use.) So $Z-X^2 - 8 = -(63/2) X^{-2} + \cdots$ and $X(Z-X^2-8) = -(63/2) X^{-1} + \cdots$ do not blow up at the puncture where $Z \approx X^2$.
At the puncture where $Z \approx - X^2$. the function $U$ blows up like $X^2$ and $V$ blows up like $X^3$.
Now, here I relied on the fact that I knew how elliptic curves work. If I have two functions $U$ and $V$ on an elliptic curve $E$, which blow up to order $2$ and $3$ at one point on the curve, and do not blow up anywhere else, then $U$ and $V$ generate the ring of functions on $E$ that only blow up at that point. Moreover, there is a relation between $U$ and $V$ of the form $V^2 + (aU+b)V + (cU^3+dU^2+eU+f)=0$. So I already knew that I had found enough functions to generate the ring, and that there would be some cubic relation between $U$ and $V$ to find.
There are some principled ways I could have found the relation, but at this point it was easier to just mess around with algebra until I worked out that $V^2 = - U^3 - 16U^2 -63 U$.
