Since $b$ is a positive integer, then $b^{2} \geq 1$, and so $a^{2} \leq 49$, which gives $1 \leq a \leq 7$, since $a$ is a positive integer. If $a=7$, then $b^{2}=50-7^{2}=1$, so $b=1$. If $a=6$, then $b^{2}=50-6^{2}=14$, which is not possible since $b$ is an integer. If $a=5$, then $b^{2}=50-5^{2}=25$, so $b=5$. If $a=4$, then $b^{2}=50-4^{2}=34$, which is not possible. If $a=3$, then $b^{2}=50-3^{2}=41$, which is not possible. If $a=2$, then $b^{2}=50-2^{2}=46$, which is not possible. If $a=1$, then $b^{2}=50-1^{2}=49$, so $b=7$. Therefore, there are 3 pairs $(a, b)$ that satisfy the equation, namely $(7,1),(5,5),(1,7)$.