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$31+15^2=256$.

Are there infinitely many solutions to:

$(2^a-1)+b^2=2^c$ with a,b,c positive integer and a,b,c different each other.

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    $\begingroup$ I mean, you can just take $b = 1$ and $a=c$. $\endgroup$
    – Marktmeister
    Commented Oct 10, 2023 at 8:20

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Yes, this has infinitely many solutions. Let $t$ be any integer, and simply take $a = t+1$, $b = 2^t-1$ and $c = 2t$. Your example is precisely the case $t=4$.

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