$$\begin{aligned} a^2 + a &= b^2 + bc \\ 4a^2 + 4a + 1 &= 4b^2 + 4bc + 1 \\ (2a + 1)^2 &= 4b^2 + 4bc + 1. \end{aligned}$$
Assume to the contrary that $c < 1$ holds. This yields
$$(2b)^2 = 4b^2 < (2a + 1)^2 = 4b^2 + 4bc + 1 < 4b^2 + 4b + 1 = (2b + 1)^2.$$
This is a contradiction as the square of an integer cannot lie strictly between two consecutive square numbers. Therefore, $c \ge 1$ holds (for instance, $a = b$ yields $c = 1$ and therefore there is a solution of the equation with $c \ge 1$).