There are precisely two available "serious" implementations of the standard algorithm for computing integral points on an elliptic curve: a non-free one in Magma (http://magma.maths.usyd.edu.au/magma/) and a free one in Sage (http://sagemath.org). The one in Sage was done by Cremona and two German masters students a few years ago, and when refereeing the Sage code, I compared the answers with Magma, and uncovered and reported numerous bugs in Magma, which were subsequently fixed. Here's how to use Sage to find all integral (or S-integral!) points on a curve over Q:
sage: E = EllipticCurve([1,2,3,4,5])
sage: E.integral_points()
[(1 : 2 : 1)]
sage: E.S_integral_points([2])
[(-103/64 : -233/512 : 1), (1 : 2 : 1)]
and here is how to use Magma:
> E := EllipticCurve([1,2,3,4,5]);
> IntegralPoints(E);
[ (1 : 2 : 1) ]
> SIntegralPoints(E, [2]);
[ (1 : 2 : 1), (-103/64 : -233/512 : 1) ]
Note that in both cases by default the points are only returned up to sign. In Sage you get both signs like this:
sage: E.integral_points(both_signs=True)
[(1 : -6 : 1), (1 : 2 : 1)]
Finally, you can use Magma for free online here: http://magma.maths.usyd.edu.au/calc/
and you can use Sage free here: https://sagecell.sagemath.org/. With Sage, you can also just download it for free and install it on your computer. With Magma, you have to pay between $100 and a few thousand dollars, depending on who you are, and deal with copy protection.
NOTE: Technically a system called SIMATH (http://tnt.math.se.tmu.ac.jp/simath/) had an implementation of computing integral points. But it was killed by our friends at Siemens Corp.
Nearly 10 years ago, I gave a talk at Wesleyan, and a gentleman named Roy Lisker asked me the same question: Fix an integral solution $(x, \ y, \ z)$ and make the substitution
$$u = 3 \ \frac {n^2 z - 12 \ x}z \qquad v = 108 \ \frac {2 \ x \ y - n \ x \ z + z^2}{z^2}$$
Then $(u, \ v)$ is a rational point on the elliptic curve $E_n: \ v^2 = u^3 + A \ u + B$ where $A = 27 \ n \ (24 - n^3)$ and $B = 54 \ (216 - 36 \ n^3 + n^6)$. (It actually turns out that $E_n$ is an elliptic curve whenever $n$ is different from 3, but I’ll discuss this case separately.)
Let me say a little about the structure of this curve for the experts:
This curve has the “obvious” rational point $T=(3 n^2, 108)$ which has order 3, considering the group structure of $E_n$. It actually turns out that these three multiples correspond to the cases $x = 0$ and $z = 0$, so if such an integral solution $(x, \ y, \ z)$ exists then the rational solution $(u, \ v)$ must correspond to a point on $E_n$ not of order 3. (Of course, I don’t care about the cyclic permutation $x \to y \to z \to x$.)
In the following table I’m computing the Mordell-Weil group of the rational points on the elliptic curve i.e. the group structure of the set of rational solutions $(u, \ v)$:
$$ \begin{matrix}
n & E_n(\mathbb Q) \\ \\
1 & Z_3 \\
2 & Z_3 \\
3 & \text{Not an elliptic curve} \\
4 & Z_3 \\
5 & Z_6 \\
6 & Z_3 \oplus \mathbb Z \\
7 & Z_3 \\
8 & Z_3 \\
9 & Z_3 \oplus \mathbb Z \\
\end{matrix} $$
Hence when $n =$ 1, 2, 4, 7 or 8 we find no integral solutions $(x, \ y, \ z)$. When $n = 5$, there are only six rational points on $E_n$, namely the multiples of $(u,v) = (3, 756)$ which all yield just one positive integral point $(x,y,z) = (2,4,1)$.
Something fascinating happens when $n = 6$... The rank is positive (the rank is actually 1) so there are infinitely many rational points $(u, \ v)$. But we must be careful: not all rational points $(u, \ v)$ yield positive integral points $(x, \ y, \ z)$. Clearly, we can scale $z$ large enough to always choose $x$ and $y$ to be integral, but we might not have $x$ and $y$ to both be positive. You’ll note that $x > 0$ if only if $u < 3 \ n^2$, so we only want rational points in a certain region of the graph. Since the rank is 1, this part of the graph is dense with rational points! Let me give some explicit numbers. The torsion part of $E_n( \mathbb Q)$ is generated by $T = (75, 108)$ and the free part is generated by $(u,v) = (-108, 2052)$. By considering various multiples of this point we get a lot of positive integral -- yet unwieldy! -- points $(x,y,z)$ such that $x/y + y/z + z/x = 6$:
$$\begin{aligned} (x,y,z) & = (12, 9, 2), \\
& = (17415354475, 90655886250, 19286662788) \\
& = (260786531732120217365431085802, 1768882504220886840084123089612, 1111094560658606608142550260961) \\
& = (64559574486549980317349907710368345747664977687333438285188, 70633079277185536037357392627802552360212921466330995726803, 313818303038935967800629401307879557072745299086647462868546) \end{aligned} $$
I’ll just mention in passing that when $n = 9$ the elliptic curve $E_n$ also has rank 1. The generator $(u,v) = (54, 4266)$ corresponds to the positive integral point $(x,y,z) = (63, 98, 12)$ on $x/y + y/z = z/x = 9$.
What about $n = 3$? The curve $E_n$ becomes $v^2 = (u – 18) (u + 9)^2$. This gives two possibilities: either $u = -9$ or $u \geq 18$. The first corresponds to $x = z$ while the second corresponds to $(z/x) \geq 4$. By cyclically permuting $x$, $y$, and $z$ we find similarly that either $x = y = z$ or $x/y + y/z + z/x \geq 6$. The latter case cannot happen by assumption so $x = y = z$ is the only possibility i.e. $(x,y,z) = (1,1,1)$ is the only solution to $x/y + y/z + z/x = 3$.
Best Answer
Finding all the integral points on an elliptic curve is a non-trivial computational problem. You say you are a "non-professional" so here is a non-professional answer: get hold of some mathematical software that does it for you (e.g. MAGMA), and then let it run until it either finds the answer or runs out of memory. Alternatively, do what perhaps you should have done at the start if you just have one curve and want to know the answer: post the equation of the curve, and hope that someone else does it for you. Here's another example of an algorithm currently used in these sorts of software (a Thue one was mentioned above but here's a different approach): find generators for the group (already computationally a bit expensive at times, depending on your luck and/or the size of sha), invoke Baker-like theorems saying "if the coordinates of the point are integral then it must be of the form sum_i n_i P_i with the n_i at most ten to the billion", and then use clever congruence techniques to massively cut down the search space by giving strong congruences for all the n_i. Then just do a brute force search.
Whether or not this will work for you, I cannot say, because it all depends on how big the coordinates of your curve are. The only clue you give so far is that the conductor is "bigger than 130000" [Edit: that was written before the OP edited the question to tell us which curve he was interested in] which of course does not preclude it being bigger than 10^10^10. Also, you need an expert to decide which of the algorithms is best for you. I'd rather do a massive amount of arithmetic in a field of tiny discriminant than a small amount of arithmetic in a field whose discriminant is so large that I can't even factor it, for example.
So in short the answer is that you're probably not going to be able to do it with pencil and paper, but there are programs around that will do it, if all you want to know is the answer.
EDIT: you posted the equation of the curve. Magma V2.15-10 says the integral points are
[ <-23, -196>, <19, 182>, <61, 784>, <-191, 28>, <103, -1442>, <-19, -144>, <-67, 592>, <23, 242>, <-49, -454>, <-157, -742>, <817, 21196>, <521, 11364>, <3857, 200404>, <10687, -910154>, <276251, -118593646> ]
plus what you get if you change all the y's to -y's.