Suppose for the sake of contradiction that there exists $k$ such that there are 2 or more. If there are at least 2 integer roots, Vieta's tells us there are 3. Let those roots be $a, b, c$. $$a + b + c = 0$$$$ab + bc + ca = -24$$$$\therefore a^2 + b^2 + c^2 = 48$$Taking modulo 4 it is easy to see $a, b, c$ must all be $\pm 4$. Contradiction to $a + b + c = 0$.

Quote: Taking modulo 4 it is easy to see $a, b, c$ must all be $\pm 4$. Why?

Here is a different approach As $k$ varies, the "naked" cubic $$y=x^3-24x$$goes up and down ("parallel" to itself, if such term could be applied to curves) along the y axis. Let us consider a point with integer coordinates on this naked cubic, for instance $x=1, y=(1)^3-24=-23$. For this point to become an integer solution of the title cubic, the whole cubic $$y=x^3-24x$$has to be shifted upwards by 23 units, to get a cubic where $k=23$, namely $$y=x^3-24x+23$$It can be seen that x=1 is a root of this equation, which thus factorizes as $$y=(x-1)(x^2+x-23)$$It can be seen that the two roots of the rightmost factor comprise a $\sqrt{3.31}$ radical detrimental to their being integers. Generalizing this finding to any point (n, n^3-24n), ones obtains the cubic $$y=x^3-24x-n^3+24n=(x-n)(x^2+nx+n^2-24)$$It can be seen that the two roots of the rightmost factor includes a $\sqrt{96-3.n^2}$ , namely a radical containing again $\sqrt{3}$ (either for real or complex roots) deleterious for them to become integers. Thus the title cubic can only have one integer root. NB I have checked n=1, 2, 3, 4 and 5 under the radical and no square turns up.

The.survivor wrote: Quote: Taking modulo 4 it is easy to see $a, b, c$ must all be $\pm 4$. Why? This is true because all values are congruent to 0 or 1 mod 4, thus in order for three of them to sum to a multiple of 4, they must all be multiples of 4. Then divide both sides by 16, to get x^2+y^2+z^2=3, this means that all are 1, or else there would be no other integer solutions. Thus a=b=c=4.