Count the ordered pairs again!

Assume that $(36a+b)(a+36b)$ is a power of 2 for some integers $a$ and $b$ . Without loss of generality we can assume $a$ and $b$ are coprime and $a<b$

Why we can assume $a$ and $b$ are coprime is because if $a$ and $b$ have a common factor $n$ then we can take this $n$ common and this $n$ would be a power of $2$ and then also $36a+b$ and $a+36b$ would also a power of $2$

Let $36a+b=2^m$ and $a+36b=2^n$

Adding and subtracting we get $37(a+b)=2^m(2^{n-m}+1)\\$ $35(b-a)=2^m(2^{n-m}-1)$

It follows that both $a+b$ and $a-b$ are divisible by $2^m$

This can happen if and only if $a$ and $b$ are divisible by $2^{m-1}$ .

Our assumption that $a$ and $b$ are coprime implies $m=1$

But then $36a+b=2$ which is impossible

Hence there are no ordered pairs that satify the given conditions.