The conjecture on odd exponents for sums of two squares

Let $a$ be a positive integer that is not a perfect power, meaning $a\ne c^k$ for every positive integer $c$ and every $k>1$. Let $m$ be the smallest positive integer such that $(a+1)/m$ is a sum of two squares. Conjecture on odd exponents.

• If $m=1$, then there are infinitely many odd positive integers $n$ such that $a^n+1$ is a sum of two squares.
• If $a\equiv 3\pmod{4}$, $a^m+1$ is a sum of two squares, and $m$ is prime, then there are infinitely many odd positive integers $n$ such that $a^n+1$ is a sum of two squares. In fact, there should be infinitely many primes $p\equiv 1\pmod{4}$ such that $a^{mp}+1$ is a sum of two squares.
• If $a\equiv 3\pmod{4}$ and $m$ is composite, then there are only finitely many odd positive integers $n$ such that $a^n+1$ is a sum of two squares.

The conjecture describes the expected split between infinite and finite families of odd exponents according to the arithmetic of $m$; the paper provides supporting results in special cases, while the general assertions remain open.