Suppose on the contrary that it does. Then $x^3+y^3+z^3=1969^2\implies x^3+y^3+z^3=4 \pmod{9}$, a contradiction, since $a^3=\pm 1,0 \pmod{9}$.

Solved with vsamc. Suppose there existed such $x,y,z$. Then we square both sides and get \[x^3+y^3+z^3=1969^2\] Note that $1969\equiv 7\pmod 9$, so $1969^2\equiv 49\equiv 4 \pmod 9$. This implies \[x^3+y^3+z^3\equiv 4\pmod 9\] Since cubes are either $-1,0,$ or $1$ mod $9$, this is not possible.

Observe that $x^3 + y^3 + z^3 \equiv -3, -2,...,2,3 \pmod{9}$, but $1969^2 \equiv 4 \pmod{9}$. QED.