Proposition 1.6 Atiyah 
Let $A$ be a ring and $\mathcal{m}$ a maximal ideal of $A$, such that every element of $1+\mathcal{m}$ is a unit in $A$. Then $A$ is a local ring.

Let $x\in A\setminus\mathcal{m}$ (It is not the quotient ring, but $A$ setminus $\mathcal{m}$). Since $m$ is maximal, the ideal generated by $x$ and $\mathcal{m}$ is $A=(1)$, that is $$(1)=(m,\{x\})$$
Edit $(m,\{x\})$ denotes the ideal generated by $\mathcal{m}$ and $x$.

Question. Why there exist $y\in A$ and $t\in\mathcal{m}$ such that $xy+t=1$?


In general, given a ring $A$, an element $x \in A$, and an ideal $I \subset A$, we have $$(\{x\}, I) = \{ax + t : a \in A,\, t \in I\}.$$
The above is easy to show. First, check that the set on the right is indeed an ideal. Second, show that any ideal containing $x$ and $I$ must contain the right set.
Apply it to your case by noting that $(\{x\}, \mathfrak{m}) = A \ni 1$.
