Integer solutions to $y^{2} = 2x^{2} +15x$.

You're looking for an $x$ such that $2x^2+15x = x(2x+15)$ is a perfect square.

Now, $\gcd(x, 2x+15) = \gcd(x,15)$, and working modulo $3$ and $5$ one quickly verifies that $x$ must be $0$ modulo $15$. So we have $x=15m$ and $$ (15m)(30m+15) $$ is a perfect square, where $3\cdot 5$ are the only prime factors in common between the two factors. Thus $m$ is itself a perfect square; so set $m=k^2$ and we're looking for $k$ such that $2k^2+1 $ is a perfect square.

From here, trial and error can relatively quickly establish that $k=2$ and $k=12$ produce solutions.

Note that $y^2=2x^2+15x$ is equivalent to $8y^2=(4x+15)^2-225$. Write $z=4x+15, 2y=w$. Then we get the equation $225= 3^2\cdot 5^2=z^2-2w^2$, so we can work in the UFD $R=\mathbb{Z}[\sqrt2]$. The fundamental unit is $u=1+\sqrt2$ of norm $-1$. We also find that $3$ and $5$ are inert in this ring. Then if $\alpha=\pm3\cdot 5(1+\sqrt2)^{2k}$ for $k\in\mathbb{Z}$, then $\alpha\tilde\alpha=225$. So: $$z=\pm\frac{\alpha+\tilde\alpha}{2}=\pm\frac{3\cdot5}{2}((1+\sqrt2)^{2k}+(1-\sqrt2)^{2k})$$ $$w=\pm\frac{\alpha-\tilde\alpha}{2}=\pm\frac{3\cdot5}{2\sqrt2}((1+\sqrt2)^{2k}-(1-\sqrt2)^{2k})$$ Now we need find when $w\equiv 0 \mod 2R$.

But $(1+\sqrt2)^{2k}=(3+2\sqrt2)^k=(3+sign(k)\cdot 2\sqrt2)^{|k|}\equiv1\mod 2R$ , so $w$ is always divisible by $2$.

Also we need to find when $z-15\equiv 0 \pmod4\iff(3+2\sqrt2)^k+(3-2\sqrt2)^k\equiv 2 \pmod8$ $\iff\sum_{i=0}^k\binom{k}{i}(2\sqrt2)^i3^{k-i}+\sum_{i=0}^k\binom{k}{i}(-1)^i(2\sqrt2)^i3^{k-i}\equiv 2\pmod8$

$$\iff2\sum_{i=0\\i \;even}^k\binom{k}{i}(2\sqrt2)^i3^{k-i}\equiv 2\pmod8$$

I haven't yet found how to exactly calculate it, but I conjecture that for $k$ even, we get infinite solutions to this equation, and they must be all the solutions.

This would give for $u=1+\sqrt2$ and $m\in\mathbb{Z}$: $$x=\pm \frac{3\cdot5}{8}(u^{4m}+\tilde u^{4m})-\frac{15}{4}$$ $$y=\pm \frac{3\cdot5}{4\sqrt2}(u^{4m}-\tilde u^{4m})$$