Why should a solution to the wave equation be finite?

It's semantics. Whoever wrote the problem prefers to refer to a wave as "A function which satisfies the wave equation and which is bounded" instead of "a function which satisfies the wave equation".

Unfortunately there are bound to be conventions which you disagree with, but in academics (undergrad and lower) the only way to deal with it is to figure out which conventions the professor (or problem writer) is working with before you read the problems. It's too easy for conversations on convention to turn into, "technically, it is a wave even though it's not physical" countered with "technically, it's not a wave because it's not bounded." The best you can do is recognize an issue in terminology ASAP and deal with it in a constructive way.

A better statement, which is more objectively true, would be: "Functions like $(x-vt)^2$ solve the wave equation, but generally don't come up and are not useful in physical solutions."

What nobody has mentioned so far is that the individual terms in the wave-function usually have a physical interpretation. For example, $\frac{\partial^2 y}{\partial t^2}$ represents an acceleration while $\frac{\partial^2 y}{\partial x^2}$ represents a force. Also, in many cases the amplitude of the wave is related to the energy density (or, in quantum mechanics, the probability density). The statements about finiteness, continuity of first derivatives, etc., all have analogues in terms of finiteness of energy, force, or the ability to localize the wave.

Also, those solutions may be valid on some finite domain that does not include singularities that are too severe.

Any function that satisfies the wave differential equation represents a wave provided that it is finite everywhere and at all times.

Has to be finite. For instance, take $f(x, t) = \frac{1}{x} + \frac{1}{t-1}$. It is not a wave, true, but it is just an example. It is not finite at all points and at all times, because at point $x=0$ we have $f(0, t) = \infty$ and at time $t = 1$ we have $f(x, 1) = \infty$. Since it is infinity, it is not finite.

$(a)$ fails when $x+vt\to\infty$.

$(b)$ fails at $x+vt \le 0$.

$(d)$ fails at $x+vt=0$