How many solutions does sin⁡(z)=10 have in the complex plane?

Just like with ${\displaystyle \sin x=\frac{1}{2}}$ there will be two supplementary angles that are solutions, that is, a pair of solutions that adds up to ${\displaystyle \pi }$. Then to each of those we can add ${\displaystyle 2\pi k}$ for integer k to handle the periodicity. That’s a countably infinite number of solutions.
These aren’t too hard to solve exactly. I’ll solve ${\displaystyle \sin z=a\text{for real}a,\left|a\right|>1}$
The expression for sine from Euler’s Formula is:
${\displaystyle \sin z=\frac{1}{2i}(e^{iz}-e^{-iz})}$
Let's let ${\displaystyle x=e^{iz}}$, then this becomes
${\displaystyle a=\frac{1}{2i}(x-\frac{1}{x})}$
${\displaystyle x^2-2aix-1=0}$
${\displaystyle x=ai\pm \sqrt{-a^2+1}}$
Since ${\displaystyle \left|a\right|>1}$ the radicand is negative, so x is purely imaginary.
${\displaystyle x=ai\pm \sqrt{-(a^2-1)}=i(a\pm \sqrt{a^2-1})=e^{i\frac{\pi }{2}}(a\pm \sqrt{a^2-1})}$
${\displaystyle \ln x=iz}$
${\displaystyle z=-i\ln x=-i(i\frac{\pi }{2}+2\pi ki+\ln (a\pm \sqrt{a^2-1}))}$
${\displaystyle z=(\frac{1}{2}+2k)\pi +i\ln (a\pm \sqrt{a^2-1})}$
The cool thing about this is that the two expressions ${\displaystyle \ln (a\pm \sqrt{a^2-1})}$ are negations of each other, as the arguments to the log are reciprocals. (I’ll leave that to you to check. It follows from the product of the x is being -1, because of the -1 in the quadratic equation.) So the z s come in conjugate pairs.
${\displaystyle z=(\frac{1}{2}+2k)\pi \pm i\ln (a+\sqrt{a^2-1})}$
Considering ${\displaystyle k=0}$ for a moment, the real part of both conjugates is ${\displaystyle \frac{\pi }{2}}$, so when they’re summed they add to ${\displaystyle \pi }$ as the imaginary parts cancel. They’re supplementary angles.
Here we have ${\displaystyle a=10}$, so
${\displaystyle \arcsin 10=(\frac{1}{2}+2k)\pi \pm i\ln (10+\sqrt{99})}$
So there are countably infinite solutions to ${\displaystyle \sin z=10}$, a conjugate pair of two solutions for each integer k.