Just some culture, in case of curiosity. If you take any prime $q$ that has an integral expression as $$ q = 2 x^2 + x y + 4 y^2, $$ such as $2, \; 5, \; 7\; 19,$ then $$ z^3 + z + 1 \equiv 0 \pmod q $$ has no roots in integral $z.$
For the special case of $31,$ we have $$ z^3 + z + 1 \equiv (z -3)(z - 14)^2 \pmod {31}. $$ For any other prime $$ p = x^2 + x y + 8 y^2, $$ such as $47, \; 67, \; 131, \; 149,$ then $$ z^3 + z + 1 \equiv 0 \pmod p $$ has three distinct roots in integral $z$ and factors as three distinct linear factors.
For any prime $r$ with $(-31|r) = -1,$ such as $3, \; 11, \; 13, \; 17,$ then $$ z^3 + z + 1 \equiv 0 \pmod r $$ has a single non-repeated root, so the cubic factors as a linear times a quadratic.
Well, why not. It turns out that every integer $n,$ positive or negative or $0,$ has an expression in integers as $$ n = x^2 + x y + 8 y^2 + z^3 + z,$$ where we deliberately strip off the 1. The difficult question is, what integers $n$ have an expression in integers as $$ n = 2 x^2 + x y + 4 y^2 + z^3 + z \; ?$$ Certainly not all, $n = \pm 1$ do not work. The first few, in absolute value, that do not work are $$ \pm 1, \; \pm 869, \; \pm 25171, \; \pm 21118439, \; \pm 611705641, $$ these being the odd integers $u$ with $$ 27 u^2 - 31 v^2 = -4. $$ The first few even values of $u$ are $$ 30 = 3^3 + 3, $$ $$ 729090 = 90^3 + 90, $$ $$ 17718345150 = 2607^3 + 2607, $$ so these are easily expressed as $2 x^2 + x y + 4 y^2 + z^3 + z $ with both $x,y = 0.$
Well, somebody did mention the discriminant of a cubic, we have $$ \mbox{disc}_z \left( z^3 + z + 1 \right) = -31 $$ and $$ \mbox{disc}_z \left( z^3 + z + u \right) = -4 - 27 u^2 = -31 v^2.$$ So there.