3. Since the right side of the equation $(x+y)^{2}=49(3 x+5 y)$ is divisible by 49, then $x+y=7 k$. Substituting this into the equation, we get $49 k^{2}=49(3 x+5 y)$ or $3 x+5 y=k^{2}$. Solving the system

$$
\left\{\begin{array}{c}
3 x+5 y=k^{2} \\
x+y=7 k
\end{array}\right.
$$

we find

$$
\left\{\begin{array}{l}
x=\frac{k(35-k)}{2} \\
y=\frac{k(k-21)}{2}
\end{array} k \in Z\right.
$$

From the obtained solutions, we need to select those for which $x$ and $y$ are integers. Since the numbers $k$ and $35-k$ and $k$ and $k-21$ have different parities, the numerators of both fractions are even for any integer $k$, and therefore, the fractions themselves are integers.
Answer: : $\left\{\begin{array}{l}x=\frac{k(35-k)}{2}, \\ y=\frac{k(k-21)}{2},\end{array} k \in Z\right.$.