Integer solutions of $x^3+y^3=z^2$ Is there any integer solution other than $(x,y,z)=(1,2,3)$ for $x^3+y^3=z^2$?

It may be of help to consider that $x^3 + y^3 = ( x + y ) \cdot ( x^2 - xy + y^2)$, so one could start by looking for values of $x$ and $y$ for which those factors are equal (and then for values where one factor "completes a square" with the other).
ADDENDUM:  I had a little time to think more on this during my snowy walk home...  We can show that if we set  $x = y$ , the two factors are $2x \cdot x^2$, which only gets us the (2, 2, 4) family of solutions [apart from $(0,0,0)$].  
In fact, if we set   $y = kx$ ($k$ integral), the factors are $[k + 1]x \cdot [k^2 - k + 1]x^2$, so we have $ k + 1 = k^2 - k + 1 \Rightarrow  k = 0, 2$ if we require the two factors $( x + y ) $ and $ ( x^2 - xy + y^2)$ to be equal.  [Just noticed that $k = 2$ will produce, for instance, the $(1,2,3)$ family.]  So almost all the solutions are ones for which  one of these factors contains a divisor just once that also appears in the other factor just once. An example is $(4,8,24)$, for which the factors are $(4+8) \cdot (4^2 - 4 \cdot 8 + 8^2)  =  ( 2^2 \cdot 3 ) \cdot (3 \cdot 4^2)$ . This ought to help narrow the search somewhat. 
