A polynomial $P(x)$ is the square of another polynomial iff the (complex) roots of $P(x)$ come in pairs. Let the roots of our product be $\alpha$, $\alpha$ and $\beta$, $\beta$ (maybe $\alpha=\beta$).
Perhaps each quadratic we see has the roots $\alpha$ and $\beta$. Since the sum of the roots is the negative of the coefficient of $x$, we must then have $k-2=-k$, that is, $k=1$. Does that work? Sure, because then $k^2=2k-1$, so the constant terms also match.
Or else the roots of the two quadratics are respectively $\alpha$, $\alpha$, and $\beta$, $\beta$. This happens iff each discriminant is $0$. So we want $(k-2)^2-4k^2=0\quad\text{and}\quad k^2-8k+4=0.\qquad\qquad(\ast)$ But $(\ast)$ forces $(-3k^2-4k+4)+3(k^2-8k+4)=0$. We get a linear equation for $k$, with integer coefficients. That would make $k$ rational. But the roots of $k^2-8k+4=0$ are not rational.
Alternately, we could find the roots of the two quadratics in $(\ast)$, by formula or by calculator, and see that the roots of one are not roots of the other.
Thus the only value of $k$ that makes our product a perfect square is $k=1$.