Simplest diophantine equation with open solvability

The perfect cuboid problem: We do not know if there is a common integer solution to $$a^2+b^2+c^2=d^2$$

$$a^2+b^2=e^2$$ $$a^2+c^2=f^2$$ $$b^2+c^2=g^2$$ with $a,b,c \ge 1$.

The last condition can also be replaced with $abc=1+h^2+i^2+j^2+k^2$.

The equations can also be reduced from four or five to one, taking $\sum(LHS-RHS)^2=0$ as a single equation in 7 or 11 variables for which we don’t know any solutions.

Determining which integers $n$ are a sum of three cubes is a very famous open problem:

$$a^3 + b^3 + c^3 = n, \quad a,b,c \in \mathbb{Z}.$$ Conjecturally, $n$ is a sum of three cubes iff $n \not \equiv 4,5 \bmod 9$.

Note that this is really a family of Diophantine equations, rather than a single Diophantine equation.

It's more complicated than the other answers by MattF and DanielLoughran, but the Erdős–Straus conjecture states that for every integer $n \ge 2$, there exist positive integers $x, y, z$ such that

$$\frac4n = \frac1x + \frac1y + \frac1z$$

Again, this is a family of Diophantine equations, related to Egyptian fractions, and computer research has verified the conjecture holds for all $n$ up to a rather large number, but whether it holds for all $n$ ($\ge 2$) is an open problem.