How would you solve the diophantine $x^4+y^4=2z^2$

I would like to know any way of solving the diophantine equation $x^4+y^4=2z^2$. Or ideas that seem worth trying out.

By solving I mean fining all solutions and proving there are no more.

Keith Conrad showed how to reduce this equation to a different one which was solved by Fermat in his notes about Fermat descent, other than I have no ideas how to solve it. I tried to do descent on it directly but that seems completely impossible so I am interested in other techniques. Thanks very much.

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page 451 of Number Theory: Analytic and modern tools by Henri Cohen shows how to reduce it to an elliptic curve. –  quanta Mar 12 '11 at 22:05

Look at pattern six: the unique coprime solution is $1^4+1^4=2 \cdot 1^2 .$
Strictly that is the unique coprime solution. So the complete solution is $(x,y,z) = (n, \pm n, \pm n^2)$ for any integer $n$ –  Henry Mar 12 '11 at 20:16