Multiplying with 4 (I first completed the square and then multiplied to get rid of fractions), the equation becomes
$$(2y+1)^2-(2x+21)^2+456=0$$
Thus
$$(2x+21)^2-(2y+1)^2=456$$
Factor the LHS
$$(2x-2y+20)(2x+2y+22)=456\\
(x-y+10)(x+y+11)=114$$
Now factor $114=2*3*19$ in all the $24$ possible ways each of them gives you exactly one solution.
P.S. The solution $x=19, y=27$ corresponds to
$$2 * 57=114$$
The solution $x=19, y=-28$ corresponds to
$$57 * 2=114$$
P.S. In general, by this method any equation of the form
\begin{align*}
y^2+ay-x^2-bx+c=0
\end{align*}
Can be reduced to an equation of teh form
\begin{align*}
(2y+a)^2-(2x+b)^2=\alpha
\end{align*}
which by factoring the LHS leads to finitely many solutions.
Equations of the form
\begin{align*}
y^2+ay-dx^2-bx+c=0
\end{align*}
can by multiplication by $4d^2$ be reduced to the Pell equation.
If you want the set of all solutions of this problem, then all you need to do, is the following:
Note that $2x+3y+4z=5 \iff 4(z+2x) + 3(y-2x) = 5$. Now, the general solution for an equation of this form is that $(z+2x,y-2x) = (-3n-1,4n+3)$, using the ordinary technique you have for two variables.
Hence, the final solution is $(x,y,z) = (k, 4n+2k+3,-3n-2k-1)$, where $n,k$ can vary among the integers.
For example, $n=13,k=-12$ gives $x = -12, y= 31, z = -16$, and $2x+3y+4z = 5$.
What helped here is the technique of reducing variables. Hence, we get the result.
Best Answer
Suppose that $x=y+n$; then $x^2-y^2=y^2+2ny+n^2-y^2=2ny+n^2=n(2y+n)$. Thus, $n$ and $2y+n$ must be complementary factors of $33$: $1$ and $33$, or $3$ and $11$. The first pair gives you $2y+1=33$, so $y=16$ and $x=y+1=17$. The second gives you $2y+3=11$, so $y=4$ and $x=y+3=7$. As a check, $17^2-16^2=289-256=33=49-16=7^2-4^2$.
If you want negative integer solutions as well, you have also the pairs $-1$ and $-33$, and $-3$ and $-11$.