<s>[INST] 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 $11^2$).
 [/INST] 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.
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