I saw in many places that the spin of photon which is a boson is 1. Which we can write as $s=1$.
I also saw that the angular momentum of a particle with spin $s$ is $\sqrt{s(s+1)}\hbar$.
If both is right, the total magnitude of the angular momentum of photon should be $\sqrt 2 \hbar$. But many resources that I found say it is just $\hbar$. I thought this was $z$-component of total angular momentum because $m_s=1, -1, 0$ for a photon.
I want to know what is right and what is wrong.
You are quite correct, but the relevant observable is $S_z$ i.e. when we measure the angular momentum we are measuring $S_z$. So it's very common to use the term angular momentum when strictly speaking we should be saying the $z$ component of the angular momentum.
One minor quibble though, massless spin one particles like photons have only the $m_s = \pm1$ states. See Why is the $S_{z} =0$ state forbidden for photons? for more on this.
In Quantum Mechanics, it is customary to speak about J angular momentum when dealing with an angular momentum whose modulus is $\sqrt{J(J+1)}\hbar$. In practice, one uses the angular momentum quantum number to characterize its values. The reason can be traced back to the passage from the old to the final form of Quantum Mechanics. In the old theory, the angular momentum quantum number was exactly proportional to the modulus.
Notice that for massless particles, it is more correct to use the helicity, i.e., the projection of the spin over the momentum, and photons may only have two values of helicity, instead of the three available to a true spin-1 particle. For more, see this link.
