Quick method /Birds eye view to determine the value 
Is there any way to guess the answer without doing elaborating calculations? 
 A: Corrected after a comment of André Nicolas:
We have $4^5=1024>1016$. This shows that $y$ can only be $1$, $2$ or $3$. Now if $y=3$, then since $3^2\times 3^5>1016$, the only possible values of $x$ are $1$ and $2$. Similarly, if $y=2$, the only possible values of $x$ are $1,2,3,4,5$. Computing the products $xy$ and comparing them with the values proposed in the question, out of these seven options there remain only two: $(x,y)=(2,2)$ and $(x,y)=(5,2)$. Checking for $z^3$, out of these two possibilities there remains only one: $(x,y,z)=(5,2,6)$.
Now if $y=1$, then $xy=x$, and all we have to check is: which of the six values among $1016-(xy)^2$ are cubes. This gives one more triple $(x,y,z)=(4,1,10)$.
So the possible values of $xy$ (among indicated) are $4$ and $10$.
A: The correct solutions are 4 and 10. Denoting $f(x,y,z) = x^2y^5 + z^3$,
x   y   z   f(x,y,z)   xy
5   2   6   1016       10
4   1   10  1016       4

Here is the PERL code used to get this:
#!/usr/bin/perl -w

use strict;

my $z = 1;
my ($y5, $z3);

while (($z3 = $z**3) < 1016) {
    my $x2y5 = 1016 - $z3;
    my $y = 1;
    while (($y5 = $y ** 5) <= $x2y5) {
        next if $x2y5 % $y5;
        my $x2 = $x2y5 / $y5;
        if (int(sqrt($x2)) ** 2 == $x2) {
            my $x = int(sqrt($x2));
            print "($x, $y, $z) -> ", ($x**2) * ($y**5) + ($z**3), "; ", $x*$y, "\n";
        }
    } continue {
        $y++;
    }
} continue {
    $z++;
}

