If $N$ is large enough, then $x^5-Nx+1$ and $x^5-Nx^2+1$ are irreducible over $\mathbb{Q}$.
For $P(x)=x^5-Nx+1$ ...
Gauss' lemma states that if a polynomial is irreducible over integers, it is also irreducible over rationals. Let's stick with integers.
Proposition 1. $P(x)=(x-\alpha_1)(x-\alpha_2)...(x-\alpha_5)$ has $4$ roots with $|\alpha_i|>1$.
Explained in the comments above, usig this source, Proposition 6. By the way, this part considers the "$N$ is large enough" criteria.
Proposition 2. $P(x)$ is irreducible over integers.
Let's suppose it's reducible, i.e. $P(x)=G(x)\cdot F(x)$, where $G, F$ are non-constant polynomials with integer coefficients. Then $P(0)=G(0)\cdot F(0)=1$. This means that the absolute values of the last coefficients of $G$ and $F$ are 1. Let's notes $$G(x)=g_kx^k+...+g_0, g_i\in \mathbb{Z}, g_k \ne 0, |g_0|=1$$ $$F(x)=f_mx^m+...+f_0, f_i\in \mathbb{Z}, f_m \ne 0, |f_0|=1$$ where $k+m=5$, $k\geq1$ and $m\geq1$
From Vieta's formulas we have that
- the absolute value of the product of $G(x)$'s roots ($G(\beta_j)=0, \beta_j \in \mathbb{C}$) is $$\left|\prod_{j=1}^k \beta_j \right|= \left|\frac{g_0}{g_k}\right|=\frac{1}{\left|g_k\right|}\leq 1 \tag{1}$$
- the absolute value of the product of $F(x)$'s roots ($F(\gamma_t)=0, \gamma_t \in \mathbb{C}$) is $$\left|\prod_{t=1}^m \gamma_t \right|= \left|\frac{f_0}{f_m}\right|=\frac{1}{\left|f_m\right|}\leq 1 \tag{2}$$
But $$\{\alpha_i\}_{i=1}^5=\{\beta_j\}_{j=1}^k \bigcup\{\gamma_t\}_{t=1}^m$$ and (from Proposition 1) only one root has $|\alpha_{i_0}|\leq1$ which falls either in $\{\beta_j\}_{j=1}^k$ or $\{\gamma_t\}_{t=1}^m$ (it can not belong to both, because $P(X)=G(X)\cdot F(X)$). But then:
- if $\alpha_{i_0}\in \{\beta_j\}_{j=1}^k$ then $\prod\limits_{t=1}^m |\gamma_t | > 1$ contradicting $(2)$
- if $\alpha_{i_0}\in \{\gamma_t\}_{t=1}^m$ then $\prod\limits_{j=1}^k |\beta_t | > 1$ contradicting $(1)$