234. $x^{2}-x y+y^{2}-x+3 y-7=0$, if it is known that the solution is: $x=3 ; y=1$.
234. Assuming $x=x_{1}+3$ and $y=y_{1}+1$, we get:

$$
x_{1}=-\frac{2 m(2 m+n)}{m^{2}-m n+n^{2}}, \quad y_{1}=-\frac{2 n(2 m+n)}{m^{2}-m n+n^{2}}
$$

The greatest common divisor (GCD) $D \left(m^{2}-m n+n^{2}, m\right)=D\left(n^{2}, m\right)=$ $= D (n, m)=1 ;$ $D \left(m^{2}-m n+n^{2}, 2 m+n\right)=$ $D \left(m^{2}-m n+n^{2}, 4 m^{2}+4 m n+n^{2}\right)=D[2 m+n, m(3 m+5 n)]=D(2 m+$ $+n, 7 m)=1$ or 7

Therefore, we need to consider the cases:

1) $m^{2}-m n+n^{2}=1$
2) $m^{2}-m n+n^{2}=2$
3) $m^{2}-m n+n^{2}=7$

Answer. $(3,1),(-1,1),(3,-1),(-3,-1),(-1,-5)$.