Solution:
The answer is yes.

The difference between the $n$th and $(n+1)$st triangular numbers is $n+1$. Therefore, a positive integer $m$ is not a triangular number if and only if it has the form
$$\frac{n^{2}+n}{2}+y$$
where $1 \leq y \leq n$. Define $x = n - y + 1$; then the conditions $1 \leq y$ and $y \leq n$ are equivalent to $y \geq 1$ and $x \geq 1$. Conversely, $n = x + y - 1$, so the polynomial
$$f(x, y) = \frac{(x + y - 1)^{2} + (x + y - 1)}{2} + y = \frac{(x + y)(x + y - 1)}{2} + y$$
hits exactly the non-triangular numbers as $x$ and $y$ range over positive integers.