I want to show that there exists some formal power series, $f(x)\in\mathbb{Z}[[x]]$, such that each consecutive partial sum is irreducible in $\mathbb{Z}[x]$. Rewording this in terms of polynomials, I want to show that there is some infinite sequence of irreducible polynomials $[p_n(x)]$ such that $\deg (p_n)\leq n$ and either $(p_{n+1}-p_n)(x) = 0$ or $(p_{n+1}-p_n)(x) = cx^{n+1}$ where $c\in\mathbb{Z}$. I can't find any examples that I can easily prove to have this property, however, I feel like there certainly ought to be such a power series. I have tried to use the polynomials in the finite fields to prove that there must be a corresponding one in the integers, but that route was not fruitful. Do there exist such non-trivial polynomial sequences?
Edit: I wish to exclude polynomial sequences that are eventually constant.
Let $q_0 \in \mathbb N$ and define $q_n$ recursively as the next prime strictly larger than $q_0+q_1+\dots+q_{n-1}$. Then the polynomial $p_n(x)=q_nx^n+q_{n-1}x^{n-1}+\dots+q_1x+q_0$ is irreducible in $\mathbb Z[x]$, and also satisfies the other conditions of the question:
$\deg p_n = n$
$(p_{n+1}-p_{n})(x) = q_{n+1}x^{n+1}\;$
What's left to prove is that $p_n(x)$ is indeed irreducible.
For that, let $P_n(z) = z^n p_n\left(\dfrac{1}{z}\right)=q_0z^n+q_1z^{n-1}+\dots+q_{n-1}z+q_n$, then $p_n$ is irreducible iff $P_n$ is, and for $|z|\le 1$:
$$ \begin{align} \left|P_n(z)\right| &\ge q_n - \left(q_0|z|^n+q_1|z|^{n-1}+\dots+q_{n-1}|z|\right) \\ &\ge q_n - \left(q_0+q_1+\dots+q_{n-1}\right) \\ &\gt 0 \end{align} $$
It follows that all roots of $P_n$ must fall outside the unit circle.
If $P_n(z)$ factored as $A(z)B(z)$, then $A(0)B(0)=P_n(0)=q_n$ and, since $q_n$ is a prime, one of the factors must be $\pm1$, which can be assumed WLOG to be $|A(0)|=1$. If $A(z)=\sum_0^k a_jz^{k-j}$ that means $|a_k|=1$, so the product of all roots of $A$ has magnitude $1/|a_0| \le 1$. But all roots of $A$ are also roots of $P_n$, which are all outside the unit circle and their product must have magnitude strictly larger than $1$. Therefore, no such factorization exists i.e. $P_n$ is irreducible, and so is $p_n$.