Actually, $\mathsf{ZF}$ easily proves that if there exists at least one non-measurable subset of $\mathbb{R}$, then there must be $2^\mathfrak{c}$ non-measurable subsets of $\mathbb{R}$.

Let $X \subseteq \mathbb{R}$ be non-measurable. Then $X = (X \cap [0,\infty)) \cup (X \cap (-\infty,0)) =: A \cup B$. If both $A$ and $B$ are measurable, then $X$ must also be measurable, at it is a finite union of measurable sets. Thus, we must have that either $A$ or $B$ is non-measurable, and we can then apply the argument in Hanul Jeon's comment.

Actually, $\mathsf{ZF}$ easily proves that if there exists at least one non-measurable subset of $\mathbb{R}$, then there must be $2^\mathfrak{c}$ non-measurable subsets of $\mathbb{R}$.

Let $X \subseteq \mathbb{R}$ be non-measurable. Then $X = (X \cap [0,\infty)) \cup (X \cap (-\infty,0)) =: A \cup B$. If both $A$ and $B$ are measurable, then $X$ must also be measurable, at it is a finite union of measurable sets. Thus, we must have that either $A$ or $B$ is non-measurable.

WLOG say $A$ is not measurable. Now there are $2^\mathfrak{c}$ measurable sets in $(-\infty,0)$, and for any of such set $C$, we have that $A \sqcup C$ is not measurable. This gives $2^\mathfrak{c}$ distinct non-measurable subsets of $\mathbb{R}$.

$\mathsf{ZF}$ easily proves that if there exists at least one non-measurable subset of $\mathbb{R}$, then there must be $2^\mathfrak{c}$ non-measurable subsets of $\mathbb{R}$.

We just need to show that there exists a non-measurable subset of some interval $(a,b)$, then apply Hanul Jeon's comment. Let $X \subseteq \mathbb{R}$ be non-measurable. Then $X = \bigsqcup_{n \in \mathbb{Z}} X \cap (n,n + 1]$. If all $X \cap (n,n+1]$ are measurable, then $X$ must also be measurable, at it is a countable union of measurable sets. Thus, there exists an $n \in \mathbb{Z}$ in which $X \cap (n,n+1] \subseteq (n,n+2)$ is not measurable.

Actually, $\mathsf{ZF}$ easily proves that if there exists at least one non-measurable subset of $\mathbb{R}$, then there must be $2^\mathfrak{c}$ non-measurable subsets of $\mathbb{R}$.

Let $X \subseteq \mathbb{R}$ be non-measurable. Then $X = (X \cap [0,\infty)) \cup (X \cap (-\infty,0)) =: A \cup B$. If both $A$ and $B$ are measurable, then $X$ must also be measurable, at it is a finite union of measurable sets. Thus, we must have that either $A$ or $B$ is non-measurable, and we can then apply the argument in Hanul Jeon's comment.