Suppose a solution exists. First note that $x$ has to be even. Indeed, if it was odd, then $y^{11}=x^2-23$ would be even but not divisible by $4$ (look mod $4$), which is impossible.
It follows that $y$ has to be odd. But then, since $y^{11}$ is congruent to $y$ modulo $4$ for odd $y$, we find that $y\equiv 1\pmod 4$.
Now comes the trick. By adding $y^{11}+2025$ to both sides, we find
$$x^2+45^2=y^{11}+2^{11}=(y+2)(y^{10}-2y^9+2^2y^8\pm\dots+2^{10})=(y+2)A.$$
Observe that A is relatively prime to $y+2$ - we have $A=y^{10}-2y^9+2^2y^8\pm\dots+2^{10}\equiv 11(-2)^{10}\pmod{y+2}$, so since $y+2$ is odd, the only common factor could be $11$, which requires $y\equiv -2\pmod{11}$. But going back to the original equation, this would imply $-1$ is a square modulo $11$, which it certainly isn't.
To sum up, we have established that $y+2$ and $A$ are relatively prime numbers whose product is a sum of two squares. However, we can see that both of them are $3\pmod 4$, so both have prime factors which are $3\pmod 4$. One of those factors must not be equal to $3$, call it $p$. By considering the displayed equation modulo $p$, we find $x^2\equiv -45^2\pmod p$, and since $p\neq 3,5$, by taking multiplicative inverse of $45$ modulo $p$ we find $z^2\equiv -1\pmod p$ has an integer solution. However, it is well-known that this is impossible whenever $p\equiv 3\pmod 4$. This gives a contradiction, showing that the equation has no solution.
(Proof inspired by the proof of Theorem 2.1 here.)
For the first: rewrite as
$$x^2 = (y^3 + 12)^2 + 48.$$
The squares differing by 48 are
\begin{align*}
(1,49) \\
(16,64) \\
(121 ,169) \\
\end{align*}
which you can get by solving $(z + 6)^2 = z^2 + 48$, $(z + 4)^2 = z^2 + 48$ and $(z + 2)^2 = z^2 + 48$ respectively.
Choosing $x$ to get the right hand squares should pose no trouble. $(y^3 + 12)^2 = 1$ has no integer solution, $(y^3 + 12)^2 = 16$ has $y = -2$ as a solution and $(y^3 + 12)^2 = 121$
has $y = -1$ as a solution.
So the solution set is
\begin{align*}
(x,y) &= (8, -2) &\lor\\
(x,y) &= (-8, -2) &\lor\\
(x,y) &= (13, -1) &\lor \\
(x,y) &= (-13, -1)
\end{align*}
Best Answer
Over 200 years ago Lagrange solved the general binary quadratic Diophatine equation
$$\rm a\ x^2 + b\ xy + c\ y^2 + d\ x + e\ y + f = 0 $$
It reduces to a Pell equation: put $\rm\ D = b^2-4ac,\ E = bd-2ae,\ F = d^2-4af\:.\ $ Then
$$\rm D\ Y^2\ =\ (D\ y + E)^2 + D\ F - E^2,\quad\quad Y\ =\ 2ax + by + d $$
Therefore if we put $\rm\quad\ \ X = D\ y + E,\quad\ \ N = E^2 - D\ F\quad\ \ $ we have the Pell equation
$$\rm X^2 - D\ Y^2\ =\ N $$
Dario Alpern has a web page Quadratic two integer variable equation solver that will solve any such equation - with complete descriptions of the methods involved. For some recent optimizations of Lagrange's algorithm see this paper H. C. Williams et al. A new look at an old equation.