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 $xy+t=1$?

 A: 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 $t=m_1y+m_2x+m_1m_2\in \mathfrak m$.
A: 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 $(\{x\}, \mathfrak{m}) = A \ni 1$.
A: Another solution I thought of: if you're familiar with $J=\operatorname{JacRad}(R)$, this is the radical who's equal to the intersection of all maximal ideals of $R$. One can easily prove that $x\in J\iff$ for every $y\in R$, $1+xy$ is a unit.
We prove $J=m$ in this particular case. The inclusion $J\subseteq m$ is straight from the definition of $J$. Now let $x\in m, y\in R$. From the definition of an ideal, $xy\in m$, and therefore $1+xy$ is a unit. This is true for all $y\in R$, so $J=m$.
Why does this imply that $R$ is a local ring? well assume for a contradiction that it's not. Then we have two maximal ideals, $m, m'$ s.t $m\not\subset m'$ and $m'\not\subset m$. By definition, $J\subseteq m\cap m'\subsetneq m$ - contradiction.
