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 $xy+t=1$?


If $a\in A$, I denote $\bar a$ the class of $a$ in $A/\mathfrak m$. Let $x\in A\setminus \mathfrak m$. Since $\mathfrak m$ is maximal, $A/\mathfrak m$ is a field. Therefore there is $\bar y\in A/\mathfrak m$ s.t. $\bar x\bar y=1$, i.e. $1=(x+m_1)(y+m_2)=xy+t$ where $t=m_1y+m_2x+m_1m_2\in \mathfrak m$.
