Greetings. I claim that the unique solution is $a=b=1$. In fact, let both the expressions $a^2+b$ and $a+b^2$ be powers of two with exponent 1 or greater. By symmetry, the highest power of two contained in $a$ could be assumed greater than or equal than that contained in $b$. In both cases dividing by a suitable power of two one can obtain an equality between odd and even numbers, Q.E.A.. Therefore $a, b$ are odd. Apart in the possibility in which $a=b=1$, by congruence considerations $a$, $b$ must leave remainder -1 on division by 4. Then put $a=4c-1$, $b=4d-1$ (f, g being positive), whence it follows (after division by 4) that both $4c^2-2c+c$ and $4d^2-2d+c$ should be powers of two with exponent greater than 1. Finally, looking at the highest power of 2 contained in the even numbers $c$, $d$ I am led again to an equality between odd and even numbers, Q.E.A..

amparvardi wrote: Find all positive integers $a$ and $b$ such that $(a + b^2)(b + a^2) = 2^m$ for some integer $m.$ (6 points) Both $a,b$ even or odd.If even,infinite descent.If odd the only solution is $(1,1)$

Let $a+b^{2}=2^{k}$ and $b+a^{2}=2^{n}$ Here $a$ and $b$ both are even or both are odd and $a+b^{2}\geq b+a^{2}\implies b^{2}-b\geq a^{2}-a\implies b\geq a$ Subtracting the two equations we get $(b-a)(a+b-1)=2^{n}(2^{k-n}-1)\implies 2^{n}|(b-a)$, since $(a+b-1)$ is odd $b-a=2^{n}p\implies b=a+2^{n}p=2^{n}-a^{2}$ $\implies a+a^{2}=2^{n}(1-p)\implies p=0\implies a=b$ Which gives $a(a+1)$ is a power of $2$ which is possible only for $a=1$ So the only possible solution is $a=b=1$

Note that $a+b^2$ and $b+a^2$ are greater than $1$ $a+b^2$ is even so $a$ and $b$ have the same parity so $a+b-1$ is odd. W.l.o.g. we can assume that $a\geq b$ then $a(a-1)\geq b(b-1)$ then $a+b^2\leq b+a^2$ Then $a+b^2|b+a^2$ then $a+b^2|(a-b)(a+b-1)$ $a+b^2$ is a power of $2$ whereas $a+b-1$ is odd so $gcd(a+b^2, a+b-1)=1$ So $a+b^2|a-b$ $0\leq a-b<a+b^2\implies a-b=0\implies a=b$ otherwise $a+b^2\not|a-b$ Then $a(a+1)=a+b^2$ is a power of $2$. $gcd(a, a+1)=1$ so atleast one of $a$ or $a+1$ is $1$ $a+1>1$ so $a=1$ so $a=b=1$ and $m=2$ Hence $(a, b, m)=(1, 1, 2)$

Another approach is like this:We can write $a=2^rs,b=2^kl,$with $s,l$ odd.setting this we find $a+b^2=2^r(s+2^{2k-r}l^2)$.Now since this is a non-negative power of $2,$$s+2^{2k-r}l^2=1\implies r=k=0,s=l=1$(similarly doing for $b+a^2$)