The diophantine eq. $x^4 +y^4 +1=z^2$

This question is an exact duplicate of the question

Does the equation $x^4+y^4+1=z^2$ have a non-trivial solution?

posted by Tito Piezas III on math.stackexchange.com.

The background of this question is this: Fermat proved that the equation, $$x^4 +y^4=z^2$$
has no solution in the positive integers. If we consider the near-miss, $$x^4 +y^4-1=z^2$$
then this has plenty (in fact, an infinity, as it can be solved by a Pell equation). But J. Cullen, by exhaustive search, found that the other near-miss, $$x^4 +y^4 +1=z^2$$
has none with $0 < x,y < 10^6$ .

Does the third equation really have none at all, or are the solutions just enormous?

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The question was also posted in January at math.stackexchange.com/questions/16887/… –  Tapio Rajala Apr 15 '11 at 9:25