The example you have chosen to start your question (water) is an atypical case. The phase diagram has a logarithmic scale for the pressure, somewhat hiding the negative slope of the transition line between the liquid and solid phases. That negative slope says that by increasing pressure, a solid in a state close to the melting curve will melt, and a liquid also close to the phase transition line will remain liquid.
However, by further increasing pressure, liquid water eventually solidifies. The reason is that the negative slope of the melting curve is connected to the lower density of normal ice Ih relative to the liquid. This is possible if the crystalline structure is open with low coordination. At high pressure, more dense solid phases with higher coordination appear, thus making a negative slope of the melting curve over the whole phase diagram unlikely.
The relation between the slope of the melting curve in the $P-T$ plane and differences in entropy $(\Delta s$ and molar volume $(\Delta v$ at the transition is ruled by Clapeyron's equation :
$$
\frac{{\mathrm d}P}{{\mathrm d}T} = \frac{s_{liq}-s_{sol}}{v_{liq}-v_{sol}}.
$$
The numerator on the right-hand side can be rewritten as $\frac{{\cal l}}{T}$, where ${\cal l}$ is the latent heat of the liquid-solid transition (positive). Therefore, the sign of $v_{liq}-v_{sol}$ controls the sign of the melting curve.
As far as I know, by increasing the pressure at a constant temperature, eventually, all materials have a fluid-crystal transition. This is consistent with the idea that by increasing pressure, the solid becomes more and more compact, making it unlikely that at very high pressure $v_{liq}-v_{sol}<0$. A final word of caution is about the possibility that, in some cases, kinetic effects may influence the practical observability of the transition to a crystalline solid.