For the Schrödinger equation $$i\hbar\frac{\partial\psi}{\partial t}=\hat H\psi({\bf x},t),$$ the Hamiltonian $$\hat H=-\frac{\hbar^2}{2m_0}\nabla^2+V({\bf x})$$ corresponds to the nonrelativistic energy-momentum relation $$\hat E=\frac{\hat p^2}{2m_0}+V({\bf x})$$ where $$\hat E=i\hbar\frac{\partial}{\partial t},\ \hat p=-i\hbar\nabla$$ So, naturally considering the relativistic energy-momentum relation \begin{equation}\label{eq:e-m}\frac{E^2}{c^2}-{\bf p}\cdot{\bf p}=m_0^2c^2\end{equation} would be the starting point to obtain a relativistic generalization of the Schrödinger equation. Replacing $E$ and ${\bf p}\cdot{\bf p}$ in \eqref{eq:e-m} by operators $$\hat E=i\hbar\frac{\partial}{\partial t}\ \mbox{and}\ \hat p\cdot\hat p=-\hbar^2\nabla^2$$ acting on a wave function $\psi$, we obtain the Klein-Gordon equation for a free particle \begin{equation}\label{eq:k-g}\left(\Box-\frac{m_0^2c^2}{\hbar^2}\right)\psi=0\end{equation} where $$\Box=-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}+\nabla^2$$
Free solutions of the Schrödinger equation with $V({\bf x})=0$ are of the form $$\psi=\exp\left[\frac{i}{\hbar}(-Et+{\bf p}\cdot{\bf x})\right]$$ They are also free solutions of the Klein-Gordon equation \eqref{eq:k-g} with the energy condition $$E=\pm c\sqrt{m_0^2c^2+p^2}$$ The solutions yielding negative energies appear to be unphysical and initially considered so by physicists, but later they were interpreted as antiparticles. Antiparticles are indeed seen in nature. In reality, antiparticles also have positive energies. Antiparticles as wave functions with negative energies is merely an interpretation of the mathematical representation of the energy condition. If antiparticles weren’t discovered, the negative energy condition would have been still thought to be unphysical.
Other than allowing solutions with negative energies, there was another issue with the Klein-Gordon equation noted by physicists. The conservation of four-current density $$j_\mu=\frac{i\hbar}{2m_0}(\psi^\ast\nabla_\mu\psi-\psi\nabla_\mu\psi^\ast),$$ where $\psi^\ast$ denotes the complex conjugate of $\psi$ and $\nabla_\mu=\left(-\frac{1}{c^2}\frac{\partial}{\partial t},\nabla\right)$, implies that the quantity $$\rho=\frac{i\hbar}{2m_0c^2}\left(\psi^\ast\frac{\partial\psi}{\partial t}-\psi\frac{\partial\psi^\ast}{\partial t}\right)$$ can be considered as a probability density. However, the problem is that $\rho$ can be negative. This is due to the appearance of first-order partial derivative $\frac{\partial\psi}{\partial t}$, which is the consequence of the Klein-Gordon equation being of second-order in time. Because of this, the Klein-Gordon equation was not regarded as a physically viable relativistic generalization of the Schrödinger equation and physicists were instead looking for a relativistic generalization of first-order in time like the Schrödinger equation. Such an equation was finally discovered by P. A. M. Dirac and is called the Dirac equation. On the other hand, the Klein-Gordon equation drew attention of physicists again after they realized that $\rho$ can be interpreted as charge density, and indeed charged pions $\pi^+$ and $\pi^-$ were discovered. Today, the Klein-Gordon equation is an important relativistic equation that describes charged spin-0 particles.
References:
[1] Walter Greiner, Relativistic Quantum Mechanics, 3rd Edition, Springer-Verlag, 2000