Is it a Proth number?
Jelly, 5 bytes
’&amp;C²&gt;

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Let \$\newcommand{\&amp;}[]{\text{ &amp; }}j\$ be a strictly positive integer. \$j + 1\$ toggles all trailing set bits of \$j\$ and the adjacent unset bit. For example, \$10011_{2} + 1 = 10100_{2}\$.
Since \${\sim}j = -(j + 1) = -j - 1\$, \$-j = {\sim}j + 1\$, so \$-n\$ applies the above to the bitwise NOT of \$j\$ (which toggles all bits), thus toggling all bits before the last \$1\$.
By taking \$j \&amp; (-j)\$ (the bitwise AND of \$j\$ and \$-j\$), all bits before and after the last set bit are nullified (since unequal in \$j\$ and \$-j\$), thus yielding the highest power of \$2\$ that divides \$j\$ evenly.
For input \$N\$, we want to apply the above to \$N - 1\$ to find \$2^{n}\$, the highest power of \$2\$ that divides \$N - 1\$. If \$m = N - 1\$, \$-m = -(N - 1) = 1 - N\$, so \$(N - 1) \&amp; (1 - N)\$ yields \$2^{n}\$.
All that's left to test is if \$2^{n} &gt; k\$. If \$k &gt; 0\$, this is true if and only if \$(2^{n})^{2} &gt; k2^{n}\$, which is true itself if and only if \$(2^{n})^{2} \ge k2^{n} + 1 = N\$.
Finally, if \$(2^{n})^{2} = N = k2^{n} + 1\$, \$2^{n}\$ must be odd (\$1\$) so the parities of both sides can match, implying that \$k = 0\$ and \$N = 1\$. In this case \$(N - 1) \&amp; (1 - N) = 0 \&amp; 0 = 0\$ and \$((N - 1) \&amp; (1 - N))^{2} = 0 &lt; 1 = N\$.
Therefore, \$((N - 1) \&amp; (1 - N))^{2} &gt; N\$ is true if and only if \$N\$ is a Proth number.
How it works
’&amp;C²&gt;  Main link. Argument: N

’      Decrement; yield N - 1.
  C    Complement; yield 1 - N.
 &amp;     Take the bitwise AND of both results.
   ²   Square the bitwise AND.
    &gt;  Compare the square to N.