Answer     4

The given equation can be rewritten as $$(x+y)^2 + 2y^2=34$$

Because both terms on the left side cannot be negative, we must have $2y^2 \le 34$ or $y^2 < 17$. This means that $-4 \le y \le 4$.

• If $y=\pm 4$, then $2y^2=32$. It is impossible.
• If $y=\pm 3$, then $2y^2=18$. It follows $(x+y)^2=16$ or $x+y=\pm 4$.
• If $y=\pm 2$, then $2y^2= 8$. It is impossible.
• If $y=\pm 1$, then $2y^2 = 2$. It is impossible.
• If $y=0$, then $2y^2=0$. It is impossible.

Therefore, we conclude there are only $\boxed{4}$ solutions: $(x, y) = (1,3)$, $(-7, 3)$, $(7,-3)$ and $(-1, -3)$.