It's actually pretty nice to reason through! This is all possible thanks to two lattice theoretic properties of Boolean rings. It's good to know about these two properties, and it turns out they're pretty instructive in this proof.
Ingredients
One thing we need is the involutory map $(x)\mapsto (1-x)$ establishing an inclusion reversing bijection on the principal ideals of the Boolean ring $R$, which I will call $L$. So $L((a)):=(1-a)$, and $(a)\subseteq (b)$ implies $L((b))\subseteq L((a))$. (Hint: if $(a)\subseteq(b)$, take $a=br$ and do something with $1-b=(1-b)(1-a)+(1-b)a$.)
The other thing afforded us in Boolean rings that we need is that $(a)\cap (b)=(ab)$. The $\supseteq$ is obvious, and I leave the $\subseteq$ as an exercise.
Using these two things, we can cook up the formula $a+b+ab$.
Searching for $(a,b)$
Let's seek the least upper bound of $(a)$ and $(b)$. A natural candidate is to find the greatest lower bound of $L((a))$ and $L((b))$, and then map it back above $(a)$ and $(b)$ using the inclusion reversing map.
Explicitly, since we know the greatest lower bound of $(1-a)$ and $(1-b)$ is $((1-a)(1-b))$ we have $(1-a)\supseteq ((1-a)(1-b))$, and after applying $L$ we have $(a)\subseteq (1-(1-a)(1-b))$, and similarly $(b)\subseteq (1-(1-a)(1-b))$.
But look at $1-(1-a)(1-b)=a+b-ab$. We've just shown that $(a+b-ab)$ is an upper bound of $(a)$ and $(b)$. Finally, it's obviously contained in $(a,b)$, so it is actually equal.
Here's a picture of that proof:
Remember that all of the subtraction signs may as well be addition because the characteristic is 2.
Summary
So in hindsight, the "mysterious" character $a+b+ab$ is just a product of the interplay of this involution $a\mapsto 1-a$ with the way principal ideals intersect. We can retrieve it via a Galois connection on the principal ideals of the ring.
Well you have the argument. Just observe that $\mathfrak{m}(M'/N)=M'/N$.
To see this observe that every element $m'\in M'$, $m'=n+a$ for some $a\in \mathfrak{m}M'$.
So $N+m'=N+n+a=N+a$. Hence $\mathfrak{m}(M'/N)=M'/N$.
Now applying the Nakayama's lemma to $M'/N$, we get $M'/N=0$.
Best Answer
Massive hint: If your ideal $I$ is generated by two elements $x$ and $y$, consider the element $ z = x + y + xy$. What is your ideal $I$ generated by now? Recall if you want to show that $(a,b) = (c)$, you need to show that $a$ and $b$ are in the right hand side, and also that $c$ is in the left hand side.
Now complete the problem by induction.