Here's a topological proof that $\mathbb{Z}$ is a PID.

Let $p,q$ be relatively prime. Then the line from the origin to the point $(p,q)\in\mathbb{R}^2$ does not pass through any lattice point, and therefore defines a simple closed curve in the torus $\mathbb{T}=\mathbb{R}^2/\mathbb{Z}^2$. Cut the torus along this curve. By classification of surfaces, the resulting surface is a cylinder. Therefore, we can reglue it to get a torus, but where our simple closed curve is now a "stupid" such thing, i.e., a ring around the torus.

Which is all to say that in this case, there exists an automorphism of the torus which takes $(p,q)\in\mathbb{Z}^2=\pi_1(\mathbb{T})$ to $(1,0)$. But this gives a matrix $\begin{bmatrix} p & x \\ q & y \end{bmatrix}\in GL_2(\mathbb{Z})$, so $py-qx\in\mathbb{Z}^{\times}$, i.e., $py-qx=\pm 1$.

The only two things this proof needs are the computation of the homology of a torus and the classification of surfaces, neither of which actually relies on $\mathbb{Z}$ being a PID!