It seems that J. Saul found a solution of {X^n + Y^n = square, n=1,2,3}.([1].Piezas) 
Smallest solution is (X,Y)=(184, 345) as follows,
184   + 345   = 23^2
184^2 + 345^2 = 391^2
184^3 + 345^3 = 6877^2

Andrew Bremner[2] showed that this equation had infinitely many solutions.

We also show this equation had infinitely many solutions and more numerical solutions.

Solution

X + Y = A^2....................................................................(1)
X^2 + Y^2 = B^2................................................................(2)
X^3 + Y^3 = C^2................................................................(3)

First, let X= px+1, Y= -px then satisfy (1).

Next, let C= qx+1 then (3) becomes to (4).

(px+1)^2-(px+1)(-px)+(-px)^2=(qx+1)^2..........................................(4)
We obtain a solution of (4), x = -(-2q+3p)/(3p^2-q^2)..........................(5)

Substitute (5) to (2), then (2) becomes to (6).

B^2=(8q^2p^2-4q^3p+q^4-12qp^3+9p^4)/((3p^2-q^2)^2).............................(6)

Hence, numerator of (6) must be square.

Set U = q/p, then V^2=U^4-4U^3+8U^2-12U+9......................................(7)

Transform (7) to Weierstrass form (8), 

M^2 = N^3-N^2-9N+9.............................................................(8)

By using Cremona's mwrank, we obtain the rank=1 and generator is P(-1, -4) on (8).

Hence, (8) has infinitely many rational solutions.


For example, the case of P(-1, -4) on (8), corresponding point is (U,V)=(12/5, -51/25) on (7).

Then we obtain (p,q)=(5, 12) and (X,Y)=(8/23, 15/23).

Accordingly, integer solution is obtained such that (X,Y)=(8/23*23^2, 15/23*23^2)=(184, 345).

Small solutions are shown below.
      [X , Y]
 P    [184, 345]
2P    [-1143896, 1562505]
3P    [184783370001360, 147916017521041]
4p    [180972102858416484216240, -49439318996063948240399]


Reference

[1]. Tito Piezas:http://sites.google.com/site/tpiezas/updates06
[2]. Andrew Bremner, Diophantine System, Internat. J. Math. & Math. Scl. Vol. 9 No. 2 (1986)