Proof

Let $I\lhd F$ be an ideal of a field $(F, + ,\cdot).$
By definition of ideals, $I$ is not empty, for all $a,b\in I$ also $a-b\in I,$ and for all $i\in I$ and all $r\in F$ we have $ir\in I$ and $ri\in I.$
It is clear from the definition of ideals, that if $I=\{0\},$ then $I$ is an ideal.
Therefore, assume, $I\neq \{0\}.$
Then, we can also choose $a\in I$ such that $a\neq 0.$
Since $F$ is a field, there is a multiplicative inverse $a^{-1}\in F.$
Therefore, $aa^{-1}\in I$ or $1\in I.$
It follows $I=F.$

Assume, ring $(R, + ,\cdot)$ has exactly two ideals.
Note that every zero ring has only one ideal.
Therefore, $R$ is not a zero ring.
Therefore, the two ideals in question of $R$ must be $\{0\}\lhd R$ and $R\lhd R,$ since these two subsets of $R$ are trivially ideals, by definition.
In particular, $1\neq 0$ holds in $R$ and there is an $a\in R$ with $a\neq 0.$
Consider the principal ideal $(a)\lhd R.$
Since $(a)\neq\{0\}$ then $1\in(a),$ therefore $1=a\cdot a^{-1}.$
We have just shown that every $a\in R$ with $a\neq 0$ has a multiplicative inverse $a^{-1}\in R.$
It follows that $R$ is a field.
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