Does the elliptic curve $y^2 = 4 x^3 -6075$ have any integer points? Let $E$ be the elliptic curve $y^2 = 4 x^3 -6075$. I ran the following  Mathematica code, which searches naively for integer solutions to $E$ but it did not find any solutions $(x, y) \in E(\mathbb{Z})$ satisfying $0 \leq x \leq 10^6$. 
T = Table[ z = 4 x^3 - 6075; 
      If[ IntegerQ[Sqrt[z]], 
         {x, Sqrt[z]}
      , 0]
  , {x, 0, 1000000}];
DeleteCases[T, 0] 

Is $E(\mathbb{Z})$ empty? Also, what is the easiest database on Mordell curves to access? I have had some trouble installing the Pari database in the past.
 A: No, $y^2=4x^3-6075$ has no integer solution. 
An elementary observation: $3\nmid x$: because $$3\mid x \implies 27\mid y^2 \implies 9\mid y \implies 
9\mid x$$ but $y^2+6075$ has no solution in $\mathbb{Z}/3^6\mathbb{Z}$. 
The ring of integer of $\mathbb{Q}(\sqrt{-3})$ is a UFD. The equation can be written as
$$\left( {\frac{{y + 45\sqrt { - 3} }}{2}} \right)\left( {\frac{{y - 45\sqrt { - 3} }}{2}} \right) = {x^3}$$
elements inside brackets, denoted by $\alpha$ and $\beta$, are integral over $\mathbb{Z}$ since $y$ is odd. 
I claim that $\alpha,\beta$ are relatively prime. If a prime $\pi$ divides both, then $\pi\mid 45\sqrt{-3}$, so $\pi = \sqrt{-3}$ or $5$. If $\pi = \sqrt{-3}$, then $3\mid x$, contradiction. If $\pi = 5$, 
let $v_5$ denote valuation at $5$, normalized so that $v_5(5)=1$, note that $v_5(\alpha) \in \mathbb{Z}$ as $v_5$ is unramified. $$0< v_5(\alpha)+v_5(\beta) = 2v_5(\alpha)= 3v_5(x) $$
this says $v_5(x)$ is even, hence $5^6 \mid (y^2+6075)$, but $y^2+6075=0$ has no solution in $\mathbb{Z}/5^6\mathbb{Z}$, ruling out $\pi = 5$. Valuation can be saved by noting that $5^3 \mid (y^2+6075)$ is already impossible, but it seems difficult to deduce the stronger $5^6 \mid (y^2+6075)$ from consideration in $\mathbb{Z}$ alone.
Since $\alpha,\beta$ are relatively prime, they are both cube, say
$$\frac{{y + 45\sqrt { - 3} }}{2} = {(\frac{{a + b\sqrt { - 3} }}{2})^3} \qquad \text{ or } \qquad \left( {\frac{{1 + \sqrt { - 3} }}{2}} \right){\left( {\frac{{a + b\sqrt { - 3} }}{2}} \right)^3}$$
with $a,b$ both even or odd. The first case gives $60= {a^2}b - {b^3}$, so $b$ has only a few possible values, checking them gives no integer solutions to original equation. The second case gives $$\frac{{45}}{2} = \frac{{{a^3} + 3{a^2}b - 9a{b^2} - 3{b^3}}}{{16}}$$ this says $v_3(a) \geq 1$, thus $v_3(a^3+3a^2b-9ab^2)\geq 3$ but $v_3(45/2) = 2$, so $v_3(3b^3) = 2$, absurd.
A: Sagemath
Multiplying by $2^4$, you get
$$
(2^2y)^2=(2^2x)^3 - 97200
$$
This is a Mordell curve with $n=-97200$, but it looks like there are only records for $-10000 \leq n \leq 10000$. The curve
$$
Y^2 = X^3-97200
$$
according to Sagemath has only one integral point
$$
(X,Y) = (49,143)
$$
so if it's correct then $(x,y) = (X/4,Y/4) = (49/4,143/4)$ is not integral and hence no integer points.  

LMFDB
The curve has conductor $6075$ so you can try to find it via the LMFDB (The L-functions and Modular Forms Database), as shown in this page. It finds up to $400000$ but lists only the minimal model of the curve so you have to match those coefficients.  
Taking modulo $2$, we know that $y$ is odd. Now let $(x,y)=(X,2Y+1)$, then
$$
\begin{align}
(2Y+1)^2 &= 4X^3- 6075\\
Y^2 + Y &= X^3 - 1519
\end{align}
$$
so the coefficients are
$$
[a_1,a_2,a_3,a_4,a_6] = [0, 0, 1, 0, -1519]
$$
corresponding to this curve in LMFDB. According to it, similarly there are no integral points.  

Magma
Another way that I know of is using an online Magma calculator

E := EllipticCurve([0, -97200]);
  IntegralPoints(E);

which returns the same result as Sagemath
[ <(49 : 143 : 1), 1> ]  

Since the curve has rank $1$, there are infinitely many rational points. Hence I think modulo $p$ methods might not work.  
Algebraic Number Theory
Perhaps another way is to solve the problem over the number field $K =\mathbb Q(\sqrt{-3})$. Rewriting, the equation becomes
$$
X^3 = (Y+180\sqrt{-3})(Y-180\sqrt{-3})
$$
Then using the fact that $K$ has class number $1$, hence unique factorization in its ring of integers, we can try to deduce the possible $X,Y$'s.  
Edit 1 Looks like this has been done by @pisco.
A: Using elementary number theory, take both sides modulo 4 so that $y^2=1\pmod{4}$, then $y=2m+1$. We also have $x=1\pmod3$ so $x=3n+1$. Expand to get:
$$4m^2+4m+1 = 4(27n^3+27n^2+9n+1)-6075$$
So
$$4m^2+4m=81n^3+81n^2+36n-6072.$$
Now take modulo $9$ on both sides to have
$$4m(m+1) = 3 \pmod9.$$
This implies that $m(m+1)=3\pmod9$. So write $m(m+1)=9a+3$ and we need to solve 
$$ m^2+m-3-9a=0. $$
It follows that $13+36a$ needs to be the square of an odd number. Maybe this helps with your numerical search?
