Let $f(x) = x^8 + x + 1$. Numerical values get large very quickly:
$$\begin{aligned} f(1) &= 3 \\ f(2) &= 259 = 7 \times 37 \\ f(3) &= 6565 = 5 \times 13 \times 101 \\ f(4) &= 65541 = 3 \times 7 \times 3121. \end{aligned}$$
These numbers may suggest that $f(n)$ will be a prime number only if $n = 1$. To prove this, we try to factorise the polynomial $x^8 + x + 1$. Progress can be made if it is suspected that $x^2 + x + 1$ is a factor. This can quickly be tested by using a cubic root of unity $\omega \neq 1$. It satisfies $\omega^2 + \omega + 1 = 0$ and $\omega^3 = 1$,
hence $\omega^8 = \omega^2$ from which we directly see $f(\omega) = 0$. Polynomial division gives now easily the factorisation $f(x) = (x^2+x+1)(x^6-x^5+x^3-x^2+1)$. Another way to obtain this factorisation is the following. We write $x^8+x+1 = x^8 - x^2 + x^2 + x + 1$ and observe
$$x^8 - x^2 = x^2(x^6 - 1) = x^2(x^3 + 1)(x^3 - 1) = x^2(x^3 + 1)(x - 1)(x^2 + x + 1).$$
This gives $f(x) = x^8 + x + 1 = (x^2 + x + 1)(x^2(x^3 + 1)(x - 1) + 1)$. If $n \ge 2$, we have $n^2 + n + 1 \ge 7$ and $n^2(n^3 + 1)(n - 1) + 1 \ge 37$, hence $f(n)$ is not a prime number if $n \ge 2$. As $f(1) = 3$ is a prime number, we conclude that $n = 1$ is the only positive integer for which $n^8 + n + 1$ is a prime number.