How to find the solution of $x^2\equiv 25\pmod{32}$?

Well, the easiest way to find a solution of $x^2\equiv 25\pmod{32}$ is to observe that $25$ is a perfect square, so $5^2=25$ which means that certainly $5^2\equiv 25\pmod{32}$ as well. If it holds in the good old integers, it certainly holds mod whatever.

You quickly earn another solution for free: $(-5)^2 \equiv 27^2 \equiv 25\pmod{32}$ as well. As you’ve observed, a simple Hensel-style lifting doesn’t work mod powers of $2$, but you can easily check that adding (or subtracting) $16$ to each of those solutions yields another one. Thus $11$ and $21$ are two other residues mod $32$ whose squares are congruent to $25$. Indeed, $(a+16)^2 \equiv a^2+32a+16^2 \equiv a^2\pmod{32}$.

Let $32|x^2-25.$ Then $x$ must be odd so let $x=2n+1.$ Then

$32|(x-5)(x+5)=(2n-4)(2n+6)\;$ iff $\;8|(n-2)(n+3).$

Now one of $n-2,n+3$ is odd and the other is even so it is necessary & sufficient that $8|n-2$ or $8|n+3.$ And the rest is easy.