Eliminating parameter from parametric equation

In Mathematica Online I tried:

Eliminate[{x == t + t^3, y == t - t^3, z == 1 + t^4}, {t}]

x^2 == -4 + y^2 + 4 z && 
x y z == -4 + 2 y^2 + 6 z - y^2 z - 2 z^2 &&
x (-2 + z) z == y (4 - 2 y^2 - 4 z - z^2) && 
x (-2 + y^2 + z) == y (2 - y^2 -3 z) && 
y^4 + y^2 (-4 + 4 z) == -4 + 8 z - 5 z^2 + z^3

Apart from the question of how to understand the answer, this is not what I expected. What I expected was

(x^2 + y^2)^2 == (x^2 - y^2)z^2

I consider my expectation reasonable because

Simplify[((t+t^3)^2 + (t-t^3)^2)^2 - ((t+t^3)^2 -(t-t^3)^2)*(1+t^4)^2] 

gives 0.

Is there a better way to achieve the elimination of t?

I also tried

Eliminate[{

  (x^2 + y^2)^2 == (x^2 - y^2)*z^2, 
  u == 2*x*(x^2 + y^2) - x*z^2, 
  v == 2*y*(x^2 + y^2) + y* z^2, 
  w == z*(y^2 - x^2)}, 
  {x, y, z}]

which gave a neat solution:

(-15u^4 - 78u^2v^2 - 15v^4)w^2 + (-48u^2 + 48v^2)w^4 + 64w^6 == 
  u^6 - 3u^4v^2 + 3u^2v^4 - v^6

Considering such a result I thought Mathematica would be able to eliminate t in my first example as well.

Edit

Just right now I discovered that the following command gives the correct and expected result:

Eliminate[{x == (t + t^3)/(1 + t^4), y == (t - t^3)/(1 + t^4)}, {t}]

Is using three parameter x, y, z instead of only x, y too much for Mathematica?