Let $k$ be a natural number . Then $3k+1$ , $4k+1$ and $6k+1$ cannot all be square numbers. 
Let $k$ be a natural number. Then $3k+1$ , $4k+1$ and $6k+1$ cannot all be square numbers.

I tried to prove this by supposing one of them is a square number and by substituting the corresponding $k$ value. But I failed to prove it.
If we ignore one term, we can make the remaining terms squares.  For example, $3k+1$ and $4k+1$ are both squares if $k=56$ (they are $13^2$ and $15^2$); $3k+1$ and $6k+1$ are both squares if $k=8$ (they are $5^2$ and $7^2$); $4k+1$ and $6k+1$ are both squares if $k=20$ (they are $9^2$ and $11^2$).
 A: I think this will work, but I don't quite have time today to fill in all details.
If $3k+1=x^2,4k+1=y^2,6k+1=z^2$, then $2x^2+z^2=3y^2$. Letting $a=x/y,b=z/y$, this gives $2a^2+b^2=3$. We can parametrize the whole space of solutions like one usually does for Pythagoream triples: $(1,1)$ is an obvious point on the curve. Now let $t$ be any real and let $Y=t(X-1)+1$ be a line through it. It intersects the curve $2a^2+b^2=3$ at point $(1,1)$ and other point $(a,b)$ satisfying $2a^2+b^2=3$ and $b=t(a-1)+1$. Substituting, we get $a^2(2+t^2)+a(2t-2t^2)+(t^2-2t-2)=0$. If we interpret this as a quadratic in $a$, $1$ is one solution, so by Viete's formulas, the other is $a=(t^2-2t-2)/(t^2+2)$. Then $b=(-t^2-4t+2)/(t^2+2)$. Now we look at $x/z=a/b=(t^2-2t-2)/(-t^2-4t+2)$. Letting $t=n/m$ be rational, this gives $x/z=(n^2-2nm-2m^2)/(-n^2-4nm+2m^2)$.
Clearly $x,z$ are relatively prime, since $2x^2-z^2=1$. I claim $n^2-2nm-2m^2,-n^2-4nm+2m^2$ have $\gcd$ dividing $6$. This is because their sum is $6nm$, and if $p\mid n$ and $p\mid n^2-2nm-2m^2$, then $p\mid m$ and vice versa.
(to be continued)
A: Assuming the contrary, set $x^2 = 3k+1$, $y^2 = 4k+1$ and $z^2 = 6k +1$
Now, has anyone tried an approach with Pell's equations?
Playing around with these three squares we get


*

*A: $4x^2 -3y^2 = 1$
which is equivalent to $u^2 - 3y^2 =1$, for $u=2x$

*B: $3y^2 - 2z^2 = 1$


and 


*

*C: $z^2 - 2x^2 = -1$


The idea is to show that there is no integer $x,y,z$ satisfying A,B and C at once. All three equations have been already studied extensively but the messy nature of the solutions of B are making it hard for me to get a contradiction.
A: This is not a complete answer, but perhaps it will help someone. If $3k = a^2-1$ and $6k = c^2-1$, then $c^2-2a^2 = -1$. It follows that the possibilities for $c$ and $a$ are the odd convergents to the continued fraction for $\sqrt{2}$ (thus $k=8$, corresponding to $c=7$ and $a=5$, is the first such value).
A: Complete answer from the literature:
I found a list of solved Pell systems in Szalay, Appendix 4, page 84. 
Kiran Kedlaya published Solving Constrained Pell Equations in Mathematics of Computation, Volume 67, April 1998, pages 833-842.
As an example of his method, on page 840 he does the harder part of the Lucas problem (bottom page 238), namely

Solving $n+1 = 2 x^2,$ $2n+1 = 3 y^2,$ $n=z^2$

with the conclusion 

Possible values of $n:$ $\{1\}$

This fits the problem above by taking $x^2 = 3k+1,$ $y^2 = 4k+1,$ $z^2 = 6k+1$ which gives the system
$$ 2 x^2 - z^2 = 1,$$
$$ 3 y^2 - 2 z^2 = 1,   $$
with the conclusion that we can only have $z^2 = 1,$ so $k=0.$
Kedlaya refers to his own solution as elementary, references bottom of page 838 to top of page 839. He mentions other articles with elementary methods; I think this is a case where "elementary" is in the eye of the beholder.
A: Edit. There must be an error in the following answer, I cannot locate it yet, will leave it as is. 
This is a reduction of the problem to another problem that I could not solve. 
Are there positive integers $t,i,j$ with $j>i$ such that: 
$\displaystyle \frac{t(t+1)}2=\frac{2i(j-i)j(j+i)}3$ ?  
I do not know if the above problem is any easier or more elementary.
The reduction to this problem is elementary (using Pythagorean triples), as follows. 
Let $x^2 = 3k+1$, $y^2 = 4k+1$, $z^2 = 6k+1$ (as in @WillJagy 's answer: I do not know if the other answers already might contain something that is equivalent to my attempt). 
Then $(x^2)^2=(3k+1)^2=(3k)^2+6k+1=(3k)^2+z^2$, hence $x^2=m^2+n^2$, $3k=2m n$, $z=m^2-n^2$ (for some integers $m>n>0$), where I used that $z$ is odd, $k$ is even. 
$1=x^2-3k=m^2+n^2-2m n=(m-n)^2$,
hence $m=n+1$, $x^2=2n^2+2n+1$, $z=2n+1$, $3k=2(n+1)n=2n^2+2n$. 
Since $x^2=m^2+n^2$, we have $x=j^2+i^2$, and
either $m=2ij, n= j^2-i^2$, or $m= j^2-i^2, n=2ij$ for some integers $j>i>0$
(the case $i=0$ resulting only in the solution  $(x,y,z)=(1,1,1)$).
Either way, $mn=(n+1)n=2ij(j^2-i^2)=2i(j-i)j(j+i)$. 
$\displaystyle z^2-y^2=2k=\frac{4(n+1)n}3$, hence (using that $z=2n+1$)
$\displaystyle y^2=4n^2+4n+1-\frac{4(n+1)n}3 =4(n+1)n(1-\frac13)+1 =$
$\displaystyle\ =\frac{8(n+1)n}3+1 =\frac{16i(j-i)j(j+i)}3+1$. Thus 
$\displaystyle y^2-1=(y-1)(y+1)=\frac{16i(j-i)j(j+i)}3$.  
Using that $y$ is odd and letting $y=2t+1$, we obtain
$\displaystyle 2t(2t+2)=4t(t+1)=\frac{16i(j-i)j(j+i)}3$, hence  
$\displaystyle \frac{t(t+1)}2=\frac{2i(j-i)j(j+i)}3$.  
Obviously the latter has no solutions with $j>i>0$ (but I do not see how to show this elementary, perhaps it is difficult, perhaps I am overlooking something) since if it did have solutions then it would produce a solution to the original problem.  It is easily seen that both sides of the latter equation are integers, indeed $\displaystyle\frac{t(t+1)}2$ is a triangular number, while if $3\not|\ i$ then $3|(j-i)j(j+i)$.  
I posted the above equation as a separate question.
