Find $(a,b)$ such that $(a^2+b)(a+b^2)=2^n$. Find all pair of positive integer numbers $(a,b)$ such that $(a^2+b)(a+b^2)=2^n$ for some positive integer number $n$.
Attempt I guest that there is only $(a,b)=(1,1)$ satisfying the assumption.
Thank you for all solution.

We assume $a\geq b$. Let $a^2+b = 2^u, b+a^2 = 2^v$ with $u\geq v\geq 1$. Then
$$a^2+b \equiv a+b^2 \equiv 0 \pmod{2^v} \implies a^4+a \equiv b^4 + b \equiv 0 \pmod{2^v}$$

If $a,b$ are both odd, we have $$a^4 + a = a(a+1)(a^2-a+1) \equiv 0 \pmod{2^v} \implies a \equiv -1 \pmod{2^v}$$ similarly $b \equiv -1 \pmod{2^v}$. Let $$a = k_1 2^v - 1 \quad \quad b = k_2 2^v - 1$$ with $k_i \geq 1$. Then
$$2^v = b+a^2 = 2^v (k_1^2 2^v - 2 k_1 + k_2) \implies 2k_1 (k_1 2^{v-1} - 1) + k_2 = 1$$
Therefore $k_2 = k_1 = v = 1$, this gives $a=b=1, n=2$. 

If $a,b$ are both even, then $$a^4 + a  \equiv 0 \pmod{2^v} \implies a \equiv 0 \pmod{2^v}$$ similarly $b \equiv 0 \pmod{2^v}$.
Letting $a= k_1 2^v, b=k_2 2^v$ gives $$2^v = b+a^2 \implies k_2 + 2^v k_1^2 = 1$$ which is not solvable.
Thus the only solution of the equation is $a=b=1$.
