Let $a,b\in\mathbb Z$ with $a\neq0$. I need to prove that if $a\nmid b$, then the equation $ax^3+bx+(b+a)=0$ does not have a solution that is a natural number.
I noticed that regardless of the values of $a$ and $b$, the equation will always have a root at $-1$ (i.e. $ax^3+bx+(b+a)=a(x^2-x+(1+\frac{b}{a}))(x+1)$.
So now the problem reduces to proving that if $a\nmid b$, $x^2-x+(1+\frac{b}{a})=0$ has no natural number solutions. I've used the quadratic formula and tried to analyse this a number of ways but I'm quite stuck.
Your quadratic equation is, with the factor of $a$ included,
$$ax^2 - ax + a + b = 0 \implies b = -a(x^2 - x + 1) \tag{1}\label{eq1A}$$
As Bill Dubuque's question comment states, for $x \in \mathbb{N}$, you have $a$ dividing the right hand side, so it must also divide the left hand side, i.e., $b$. Thus, you require $a \mid b$. However, since you're given $a \not\mid b$, this can't be true, meaning there is no natural number solution for $x$.