Starting from the given equation, we obtain the equivalent equations $5x^{2}-4xy+2x+y^{2}=624$. Adding 1 to both sides, we have $5x^{2}-4xy+2x+y^{2}+1=625$. Rewriting, we get $4x^{2}-4xy+y^{2}+x^{2}+2x+1=625$. Completing the square, we have $(2x-y)^{2}+(x+1)^{2}=625$. Note that $625=25^{2}$. Since $x$ and $y$ are both integers, then the left side of the given equation is the sum of two perfect squares. Since any perfect square is non-negative, then each of these perfect squares is at most $625=25^{2}$. The pairs of perfect squares from this list that have a sum of 625 are $625=625+0=576+49=400+225$. Therefore, $(2x-y)^{2}$ and $(x+1)^{2}$ equal $25^{2}$ and $0^{2}$ in some order, or $24^{2}$ and $7^{2}$ in some order, or $20^{2}$ and $15^{2}$ in some order. Thus, $2x-y$ and $x+1$ equal $\pm 25$ and 0 in some order, or $\pm 24$ and $\pm 7$ in some order, or $\pm 20$ and $\pm 15$ in some order. Since $x \geq 0$, then $x+1 \geq 1$, so we need to consider the possibilities that $x+1=25,24,7,20,15$: - If $x+1=25$, then $x=24$. If $2x-y=0$ and $x=24$, then $y=48$. - If $x+1=24$, then $x=23$. If $2x-y=7$ and $x=23$, then $y=39$; if $2x-y=-7$ and $x=23$, then $y=53$. - If $x+1=7$, then $x=6$. If $2x-y=24$ and $x=6$, then $y=-12$; if $2x-y=-24$ and $x=6$, then $y=36$. - If $x+1=20$, then $x=19$. If $2x-y=15$ and $x=19$, then $y=23$; if $2x-y=-15$ and $x=19$, then $y=53$. - If $x+1=15$, then $x=14$. If $2x-y=20$ and $x=14$, then $y=8$; if $2x-y=-20$ and $x=14$, then $y=48$. From this list, the pairs of non-negative integers $(x, y)$ that satisfy the condition $0 \leq x \leq y$ are $(x, y)=(24,48),(23,39),(23,53),(6,36),(19,23),(19,53),(14,48)$. There are 7 such pairs.