7

For the equation

$$x^3+y^3+39xy=2025$$

We can re-write as

$$\begin{array}{rl}(x+y)(x^2-xy+y^2)+39xy& =2025\\ (x+y)((x+y{)}^2-3xy)+39xy& =2025\end{array}$$

Let

$$a=x+y,b=xy$$

We can re-write the equation as

$$\begin{array}{rl}a(a^2-3b)+39b& =2025\\ a^3-2025& =3b(a-13)\end{array}$$

Let $c=a-13$，then $a=c+13$. So we have

$$a^3=(c+13{)}^3=c^3+39c^2+507c+2197$$

Then we have

$$a^3-2025=(c^3+39c^2+507c+2197)-2025=c^3+39c^2+507c+172$$

We know that $3b(a-13)=3bc$, so

$$c^3+39c^2+507c+172=3bc$$

While $c\ne 0$, we have

$$3b=\frac{c^3+39c^2+507c+172}{c}=c^2+39c+507+\frac{172}{c}$$

We should make $3b$ an integer, So we should make sure that $\frac{172}{c}$ is an integer, which is $c\mid 8$
We can see $172=2^2\times 43$, all the factors are
$\pm 1,\pm 2,\pm 4,\pm 43,\pm 86,\pm 172$
We can make another equation to see if $x$ and $y$ have integer solutions
Consider the following equation

$$t^2-at+b=0$$

We know the discriminant

$$\Delta =a^2-4b=(c+13{)}^2-4b=c^2+26c+169-4b\ge 0$$

Substitute, we have

$$c^2+26c+169-4b=c^2+26c+169-\frac{4(c^3+39c^2+507c+172)}{3c}$$ $$=\frac{-c^3-78c^2-1521c-688}{3c}$$

Substitute all the factors into the equation, only

$$c=-43$$

makes the discriminant equals to 0

$$a=-43+13=-30$$

We have the solution

$$t=\frac{a}{2}=-15$$

So we can get

$$\boxed{\{\begin{array}{rl}& x=-15\\ & y=-15\end{array}}$$