Example 3 Find all triangles with integer side lengths and an incircle radius of 1.

Find all triangles with integer side lengths and an incircle radius of 1.
Let the three sides of the triangle be $a$, $b$, and $c$, with $a \geqslant b \geqslant c$, and let $p=\frac{1}{2}(a+b+c)$. By Heron's formula and the area formula, we have
$$
p \times 1=\sqrt{p(p-a)(p-b)(p-c)},
$$

which simplifies to $p=(p-a)(p-b)(p-c)$.
Thus, $4(a+b+c)$
$$
=(b+c-a)(c+a-b)(a+b-c) \text {. }
$$

Therefore, among the three factors on the right side of equation (1), one must be even, and the three factors must have the same parity. Hence, we can let
$$
\begin{array}{l}
x=\frac{1}{2}(b+c-a), \\
y=\frac{1}{2}(c+a-b), \\
z=\frac{1}{2}(a+b-c) .
\end{array}
$$

Then $x \leqslant y \leqslant z$, and $x$, $y$, and $z$ are positive integers, and
$$
x+y+z=x y z \text {. }
$$

Clearly, $x+y+z \leqslant 3 z$.
Thus, $x y \leqslant 3$.
If $x y=3$, from equation (2) we get $x=y=z$, which contradicts the fact that $x$ and $y$ are integers.

If $x y=2$, then $x=1$, $y=2$, and from equation (2) we get $z=3$.

If $x y=1$, then $x=1$, $y=1$, and equation (2) becomes $z+2=z$, which is a contradiction.
Therefore, equation (2) has only one integer solution
$$
x=1, y=2, z=3 \text {. }
$$