A Complex-Analytic Proof
We shall first verify that exactly one root $\zeta\in\mathbb{C}$ of $f(x):=x^5+2x+1$ satisfies $|\zeta|<1$. Let $\epsilon$ be an arbitrary real number such that $0<\epsilon<\frac{1}{4}$. Consider the open ball $B_{1-\epsilon}(0)\subseteq \mathbb{C}$ centered at $0$ with radius $1-\epsilon$. We see that, for a complex number $z$ on the boundary $\partial B_{1-\epsilon}(0)$ of $B_{1-\epsilon}(0)$,
$$|2z|=2(1-\epsilon)\text{ and }|z^5+1|\leq |z|^5+1=(1-\epsilon)^5+1\,.$$
Observe that
$$\begin{align}(1-\epsilon)^5&=1-5\epsilon+\epsilon^2(10-10\epsilon+5\epsilon^2-\epsilon^3)
\\
&< 1-5\epsilon+\epsilon^2\left(10-0+\frac{5}{4^2}-0\right)
\\
&<1-5\epsilon+11\epsilon^2\,,\end{align}$$
as $0<\epsilon<\frac14$.
That is, $3-11\epsilon>0$, and so
$$(1-\epsilon)^5<1-5\epsilon+11\epsilon^2=(1-2\epsilon)-\epsilon(3-11\epsilon)<1-2\epsilon\,.$$
Consequently,
$$|2z|>\left|z^5+1\right|$$
for $z\in\partial B_{1-\epsilon}(0)$. By Rouché's Theorem, the number of roots of $$f(z)=z^5+2z+1=(2z)+\left(z^5+1\right)$$ in $B_{1-\epsilon}(0)$ is the same as the number of roots of $2z$ in $B_{1-\epsilon}(0)$, which is $1$. Hence, $f(z)$ has exactly one root $z=\zeta$ inside $\bigcup\limits_{\epsilon\in\left(0,\frac{1}{4}\right)}\,B_{1-\epsilon}(0)=B_1(0)$.
It is also easy to see that $f(x)$ has no root of modulus $1$. This is because $f(r)=0$ with $|r|=1$ implies
$$1=|r|^5=\big|r^5\big|=|-2r-1|\geq 2|-r|-|-1|=2|r|-1=2\cdot 1-1=1\,,$$
whence we have an equality. By the equality condition of the Triangle Inequality, we must have $r=-1$, but $f(-1)=-2\neq 0$. Ergo, the roots of $f(z)$ other than $z=\zeta$ have moduli larger than $1$.
Finally, if $f(x)=p(x)\,q(x)$ for some nonconstant $p(x),q(x)\in\mathbb{Z}[x]$, then we can assume without loss of generality that $p(\zeta)=0$. Ergo, all the roots of $q(x)$ must have moduli greater than $1$. That is, the constant term of $q(x)$ must be an integer $c$ with $|c|>1$. However, $c$ must divide the constant term of $f(x)$, which is $1$. This is absurd. Therefore, $f(x)$ is an irreducible polynomial over $\mathbb{Z}$ (whence also over $\mathbb{Q}$).