Set of quadratic expressions $nx^2+m$ whose union is all integers?

Proposition: Let $c\ne 0$ be any integer. Then there exists a multiplier $B\ge 1$ such that $(Bn)^2+c$ is never a square (inclusive of $0$) for any $n\ge 1$.

Proof: There are only finitely many ways to write $c$ as the product of two integers. Choose $B$ so that $2B$ exceeds the maximum difference between any complementary factors of $c$. Then $(m-Bn)(m+Bn) = c$ has no integer solutions with $n\ge 1$.

Lemma: Let $A$ be any finite set of integers, and $c \ne A$. There exists a $B\ge 1$ such that the set $\{(Bn)^2+c: n > 0\}$ is disjoint from the union of quadratic progressions $\{ n^2 + a : n \ge 0, a \in A \}$.

Proof: Apply the proposition to each value $c-a$, and take $B$ to be the LCM of all the (finitely many) multipliers so obtained.

Theorem: There exists an infinite sequence $(B_k, c_k) : k \ge 0$ with $B_k > 0$ such that the quadratic progressions $\{ B_k n^2 + c_k : n \ge 0 \}$ form a partition of $\mathbb N$.

Proof: Start with $(B_0,c_0) = (1,1)$. We proceed inductively: suppose that $(B_k, c_k)$ have been chosen for all $k<m$. By Steven Stadnicki's comment above, there exists a least element of $\mathbb N$ not yet covered: call it $c$. Necessarily, $c \ne c_k$ for any $k<m$. Now choose $c_m = c$ and $B_m$ according to the lemma (using $A = \{c_0,\ldots,c_{m-1}\}$). By construction $\{ B_m n^2 + c_m : n > 0 \}$ is disjoint from all previous progressions, and also $\{ B_m n^2 + c_m : n = 0 \}$ is disjoint by choice of $c$. Thus we may construct an infinite sequence $(B_k,c_k)$ in this manner. Finally, this is certain to cover all of $\mathbb N$ since we chose $c$ minimally, so that the first $m$ progressions necessarily cover $\{1,\ldots,m\}$.