If you want to use a contradiction:
If $x \ne y$ then $x^3+x \ne y^3+y$
EDIT: this is true because $x^3+x$ is an increasing function (its derivative is $3x^2+1$, which is always > 0). Therefore:
if $x>y$, $f(x)>f(y)$ ==> $x^3+x$ > $y^3+y$
if $x<y$, $f(x)<f(y)$ ==> $x^3+x$ < $y^3+y$
Some said the "integer" requirement is redundant. Well, actually it isn't. It's just superabundant.
In fact, that statement is true only if $x$ and $y$ are real numbers (integers are included in the real field). If there were no restrictions, we would have to consider $x$ and $y$ as complex numbers, and this statement would not be true.
In the case of my proof, $x^3 + x=y^3 + y$ would not imply $x=y$ because of the Fundamental Theorem of Algebra. Considering $y$ as a constant in $x^3 + x=y^3 + y$ (which can be factored in $(x-y)(x^2+xy+y^2+1) = 0$), we would obtain 3 solutions. One being real ($x=y$) and two being complex conjugate:
You can see why, if $x$ and $y$ are complex, $x^3 + x=y^3 + y$ does not imply $x=y$.