From $$\zeta (s)=\sum_{n\ge 1}n^{-s}\quad\Re (s)\gt 1$$ we know that $$\zeta (s)\ne 0\quad\Re (s)\gt 1$$ and from the functional equation for $\zeta (s)$ that $$\zeta (s)\ne 0,\,\Re (s)\lt 0,\, s\ne -2n,\,n\in\mathbb N.$$ We can also prove $$\zeta (s)\ne 0\quad\Re (s)=1, \tag*{(*)}$$ possibly using the prime number theorem.
Now, this all proves that if $$\zeta (s)=0,\,s\ne -2n,\,n\in\mathbb N, \qquad (\dagger )$$ then all the non-trivial zeros lie on the strip $0\lt\Re (s)\lt 1$. I found numerous proofs of $(*)$, but no proof of $(\dagger)$. The presence of all non-trivial zeros on the critical strip allegedly follows immediately from $(*)$, but I think it does not, since one must rule out the case that $\zeta (s)$ has no zeros except for negative even integers.
Hadamard's theorem implies that the entire function $\xi(s)=(s-1)\pi^{-s/2} \Gamma(s/2) \zeta(s)$ is of exponential order $1$ and maximal type (see the behaviour on the real axis and Stirling formula for maximal type) so it has roughly $R\log R$ roots in a circle of radius $R$ and the sum on the reciprocal absolute value of the roots diverges; since the roots of $\xi$ are the non-trivial roots of $\zeta$ the result (and the correct order of the number of non trivial roots with imaginary part up to $T$ conjectured by Riemann and proved by Von Mangoldt) follow from basic entire function theory
The Euler product of $\zeta$, non-vanishing of $\Gamma= \frac{1}{\Delta}, \Delta$ entire and the functional equation $\xi(s)=\xi(1-s)$ immediately imply that all the zeros must lie in the (closed) critical strip from simple classical complex analysis and the functional equation (however the proof that there are no zeroes on the line $\Re s=1$ is quite hard)